Chapter 31 The Law of Refraction

$Page 1: Chapter 31 The Law of Refraction$
Chapter 31 The Law of Refraction“If in other sciences we should arrive at certaintywithout doubt and truth without error, it behooves usto place the foundations of knowledge inmathematics.” Roger Bacon

31.1 RefractionIn chapter 30, we saw that an incident ray of light is reflected from a piece ofreflecting material such that the reflected ray makes the same angle with thenormal as the incident ray. If the reflecting surface is a boundary between twodifferent transparent mediums, such as air and glass, some of the incident light isalso transmitted into the glass, as shown in figure 31.1. However, it is observed

Figure 31.1 Reflection and refraction of light.

experimentally that this transmitted ray of light is bent as it enters the secondmedium. The bending of light as it passes from one medium into another is calledrefraction. Refraction of light occurs because light travels at different speeds indifferent mediums. Light traveling through a vacuum travels at the speed c = 3.00 ×108 m/s. But when light enters a medium there is a complex interaction between theelectromagnetic wave (light) and the atomic configuration of the medium. Thisinteraction causes the electromagnetic wave to slow down in the medium. Thisslowing down of the wave as it goes from a vacuum into the medium causes it tobend. We will use Huygens’ principle to show how this is accomplished in section31.2.

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31.2 The Law of RefractionLet us consider a wave front B1B2 of a plane parallel monochromatic waveimpinging on the boundary of two different mediums, as shown in figure 31.2. Theincident ray makes an angle of incidence i with the normal N. The incident light

Figure 31.2 The law of refraction by Huygens’ principle.

moves at a speed v1 in medium 1 and v2 in medium 2, and we assume that v1 isgreater than v2. The incident wave has just touched the boundary at B1. In a time∆t, B2, the upper portion of the initial wave front, travels a distance v1∆t, andimpinges at the boundary of the interface at B’2.

In this same time interval ∆t, the wave front at B1 enters the second medium.By Huygens’ principle, a secondary wavelet can be drawn emanating from the pointB1. This wave moves a radial distance v2∆t in the second medium in the timeinterval ∆t, and is shown as the circle of radius v2∆t in the figure. The radialdistance v2∆t is less than the distance v1∆t because v2 is less than v1. By Huygens’principle, the line drawn from B’2 that is tangent to the secondary wavelet is thenew wave front. The point of tangency is denoted by B’1 and the new wave front inmedium 2 is B’1B’2 .The radius from B1 to B’1, when extended, becomes the refractedray B1C. The other refracted rays are drawn parallel to B1C, as shown in figure 31.2.The angle that the refracted ray makes with the normal is called the angle ofrefraction r.

We can obtain the relation between the angles i and r from the geometry offigure 31.2. Since line B2B’2 makes an angle i with the dashed normal, angle B2B’2B1

is equal to (900 − i), and since the sum of the angles in triangle B1B2B’2 must equal1800, it follows that angle B2B1B’2 is also equal to the angle i. Using similarreasoning, angle B1B’2B’1 is equal to r. Hence, from the trigonometry of figure 31.2,

Chapter 31 The Law of Refraction

31-2

sin i = v1∆t (31.1) B1B’2

andsin r = v2∆t (31.2) B1B’2

Let us divide equation 31.1 by equation 31.2 and obtain

sin isin r =

v1✁tB1B2

∏

v2✁tB1B2

∏

and (31.3)sin i

sin r = v1v2 = constant = n21

Equation 31.3 is the law of refraction. It says that the ratio of the sine of the angleof incidence to the sine of the angle of refraction is equal to the ratio of the speed oflight in medium 1 to the speed of light in medium 2. Because the speed of light inmedium 1 v1 is a constant and the speed of light in medium 2 v2 is a constant, thentheir ratio v1/v2, must also be a constant. This constant is called the index ofrefraction of medium 2 with respect to medium 1 and is denoted by n21.

If medium 1 is a vacuum, then v1 = c, and the index of refraction of themedium with respect to a vacuum is

n = c (31.4) v

Since the speed v in any medium is always less than c, the index of refraction, n =c/v, is always greater than 1, except for in a vacuum where it is equal to 1. Indices ofrefraction for various substances are given in table 31.1. Notice that the index ofrefraction of air is so close to the value 1, the index of refraction of a vacuum, that inmany practical situations, air is used in place of a vacuum.

The law of refraction can be put in a more convenient form by using equation31.4. We can write the index of refraction of medium 1 with respect to a vacuum as

n1 = c (31.5) v1

whereas we can write the index of refraction of medium 2 with respect to a vacuumas

n2 = c (31.6) v2


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1.000291.52.421.521.57-1.721.461.471.311.491.541.33

AirBenzeneDiamondGlass, crownGlass, flintGlass, fused quartzGlycerineIcePlexiglassQuartz crystalWater

nSubstance

Table 31.1 Index of Refraction for Various Materials (λ = 589.2 nm, the D line of sodium)

Solving for the speeds v1 and v2 from equations 31.5 and 31.6, respectively, andsubstituting them into equation 31.3, gives

n21 = v1 = c/n1 = n2 (31.7) v2 c/n2 n1

Using equation 31.7, we can write the law of refraction, equation 31.3, as

sin i = n21 = n2 sin r n1

orn1 sin i = n2 sin r (31.8)

Equation 31.8 is the form of the law of refraction that we will use in what follows. Itis also called Snell’s law after its discoverer, Willebrord Snell (1591-1626) a Dutchmathematician who discovered it in 1620, the same year the Pilgrims landed atPlymouth Rock. Note that if a ray lies along the normal, then the angle of incidencei is equal to zero, and hence the angle of refraction r must also be zero, and there isno refraction of this ray.

The fact that the speed of light varies from medium to medium has animportant effect on the wavelength of light. When an initial wave enters a secondmedium, its wavelength changes. We can see this from equation 31.3, where

n21 = v1 v2

However, the speed of any wave is given by


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v = λν (31.9)

where λ is the wavelength and ν is the frequency of the wave. Hence,

n21 = v1 = λ1ν = λ1 (31.10) v2 λ2ν λ2

The frequency ν of the wave does not change as it goes across the boundary becausethere are the same number of wave fronts passing from medium 1 into medium 2, perunit time1. Therefore,

λ2 = λ1 = λ1 = n1 λ1 (31.11) n21 n2/n1 n2

That is, the wavelength of the light in the second medium λ2 is less than the initialwavelength λ1 by the factor 1/n21 = n1/n2. To summarize, the speed of light variesfrom one medium to another but the frequency of the light remains the same in bothmediums. Because of the changing speed of light, the wavelength of the light changesas it goes into the second medium.

Example 31.1

The refraction of light. A ray of light of 500.0 nm wavelength in air, impinges on apiece of crown glass at an angle of incidence of 35.00. Find (a) the angle of refraction,(b) the speed of light in the glass, and (c) the wavelength of light in the glass.

Solution

a. We find the angle of refraction from Snell’s law, equation 31.8, with the indices ofrefraction n1 = nair = 1.00 and n2 = nglass = 1.52, found in table 31.1. Therefore,

n1 sin i = n2 sin rsin r = n1 sin i

n2

= 1.00 sin 35.00 = 0.3771.52

r = sin−1 0.377and the angle of refraction is

r = 22.20


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1That is, you can’t have 5 waves per second incident on the boundary between the two mediums andonly 3 waves per second transmitted through the boundary. Where did the other waves go? You musthave the same number leaving the boundary as you have entering the boundary. Hence thefrequency of the wave must be the same on both sides of the boundary.

b. The speed of light in the glass, found from equation 31.6, is

v2 = c = 3.00 × 108 m/s n2 1.52

= 1.97 × 108 m/s

c. The wavelength of light in the glass, found from equation 31.11, is

λ2 = n1 λ1 n2

= (1.00) (500.0 nm)1.52

= 329 nm

To go to this Interactive Example click on this sentence.

The law of refraction has been derived by Huygens’ principle, treating lightas a wave phenomenon. We will now simplify the analysis of refraction by usingonly the ray model of light. Thus, the refraction of waves in figure 31.2 will now beshown as the equivalent refraction of rays in figure 31.3. The dashed line in figure31.3(a) is the direction that the incident ray would have followed if it had not

Figure 31.3 The law of refraction by a ray diagram.

entered the second medium. Note that the incident ray was bent away from itsoriginal direction and bent toward the normal. Medium 1 has an index of refractionthat is less than medium 2. The medium with the smaller value of the index ofrefraction is called the rarer medium, whereas the medium with the larger value ofn is called the denser medium. Hence, whenever a ray of light goes from a rarer


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http://www.farmingdale.edu/faculty/peter-nolan/pdf/univphys/UPEX31-01.XLS

medium to a denser medium the refracted ray is always bent toward the normal. Thelarger the value of the index of refraction n2, the greater the amount of bending.

By the principle of reversibility, a light ray that reverses the path in figure31.3(a), goes from a denser medium to a rarer medium and is bent away from thenormal, as seen in figure 31.3(b). Hence, whenever a ray of light goes from a densermedium to a rarer medium, the refracted ray is bent away from the normal.

31.3 Apparent Depth of an Object Immersed in WaterAn interesting example of the refraction of light is the observation of an object whenit is under water. A stick immersed in water appears to be bent, figure 31.4(a), anda fish in water is not where it seems to be. Sometimes in a carnival or a bazaar, agame is played where a glass is placed in water, as in figure 31.4(b), and the

(a) (b)Figure 31.4 (a) The stick appears bent because of refraction at the water’s surface.

(b) The apparent depth of an object immersed in water.

patrons try to throw a quarter into the glass, receiving a prize if they are successful.Of course it is much more difficult than it appears because the glass is not where itseems to be.

As an example, let the object, the quarter in the glass, be at the bottom of thewater at Q, a distance p from the top of the water. A ray from Q makes an angle ofincidence i with the normal. Because the ray is going from the more dense medium,water, to the less dense medium, air, the refracted ray is bent away from thenormal, as shown. The observer’s eye, located at R, sees the ray ROQ, but theperson believes that light rays travel in straight lines, and thinks he sees the rayROI. The person, therefore, assumes that the quarter in the glass is located at I, adistance q below the surface of the water. In a sense, you might say that he is


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looking at the image of the glass. The vertical ray QN’ lies along the normal andhence is not refracted, as shown. These two rays QOR and QN’ do not intersectanywhere in medium 2, the air, but when produced backward, the ray ROIintersects QN’ at I, and in this sense, I appears as the image of Q. The glass appearsto be much closer to the top of the water than it really is. The distance q from thetop of the water to I is called the apparent depth of the object. Let us now solve forthe apparent depth.

From Snell’s law of refraction, equation 31.8, we have

n1 sin i = n2 sin r

We determine the angles i and r from the geometry of figure 31.4 by observing that

tan i = x (31.12) p

andtan r = x (31.13) q

To simplify the solution, let us assume that the angles i and r are small enough (lessthan 100) to use the small-angle approximation, that is,

sin θ ≈ tan θ

With this assumption, equations 31.12 and 31.13 can be substituted back intoequation 31.8 as

n1 tan i = n2 tan rn1 x = n2 x

p qSolving for the apparent depth,

q = n2 p (31.14) n1

Example 31.2

Apparent depth. A glass is placed in the bottom of a tub of water 25.0 cm deep. Whatis its apparent depth?

Solution

The apparent depth, found from equation 31.14, is


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q = n2 p n1

= (25.0 cm)11.33= 18.8 cm

The glass appears to be 18.8 cm below the top of the water when in fact it is really25.0 cm below the top.


In the solution of the problem of the apparent depth, we made anassumption, just as is often done in physics. The solution is, of course, only as goodas the assumption, that is, that the angles i and r are less than 100. Suppose theyare not. What effect does this have on the solution? For example, if x in figure 31.4is 12.5 cm then the angle i, found from equation 31.12, is

i = tan−1 x = tan−1 12.5 p 25.0

= 26.60

The angle r, found from Snell’s law, is

sin r = n1 sin i n2

= 1.33 sin 26.60

1.00 and the angle of refraction is

r = 36.50

This is obviously much larger than the assumed limit of 100. The apparent depth,now found from equation 31.13, is

q = x = 12.5 cm tan r tan 36.50

= 16.9 cm

If the calculation is repeated with x = 25.0 cm, the angle of incidence i is now equalto 45.00, and the angle of refraction r is now equal to 70.10 and the apparent depth qis 9.05 cm. These values are significantly different from the values obtained withthe initial assumptions. This is again a reminder that in trying to represent thephysical world in terms of mathematical equations, certain assumptions are made.


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When these assumptions are valid, the equations are good. When they are not validthe equations no longer represent the physical world and are essentially useless.

31.4 Refraction through Parallel FacesA ray of light passing through two parallel boundaries is refracted at each of theseboundaries, but the final ray is in the same direction as the initial ray, as shown infigure 31.5. An incident ray makes an angle i1 with the normal N. Because medium

Figure 31.5 Refraction through parallel faces.

2 is more dense than medium 1, the refracted ray is bent toward the normal inmedium 2, making an angle of refraction of r2 with the normal. Snell’s law for thefirst interface becomes

n1 sin i1 = n2 sin r2 (31.15)

This refracted ray now becomes the incident ray for the interface between medium 2and medium 1, and makes an angle i2 with the normal N’ .Since normals N and N’are parallel, angle r2 is equal to angle i2. As the ray goes from the more densemedium 2 to the less dense medium 1, the ray is refracted away from the normalthrough the angle r1, as shown. Snell’s law for the refraction at this interface is

n2 sin i2 = n1 sin r1 (31.16)

Because angle r2 is equal to angle i2, the right-hand side of equation 31.15 is equalto the left-hand side of equation 31.16. Equating them, we obtain


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n1 sin i1 = n1 sin r1

and i1 = r1

Thus, the angle r1 is equal to the initial angle i1, and the final refracted ray comesout parallel to the direction of the original ray, but slightly displaced.

Example 31.3

Refraction through parallel faces. A ray of light in air makes an angle of incidence of40.00 with the normal to a plate of glass of n = 1.50, as in figure 31.5. Find the angleof refraction of the ray in the glass and the angle of refraction of the final ray as itpasses from the glass back into the air.

Solution

The angle of refraction for the first interface, found from Snell’s law, is


sin r2 = n1 sin i1

n2

= 1.00 sin 40.00 = 0.4285 1.50

and the angle of refraction for the first interface becomes

r2 = sin−1 (0.4285)r2 = 25.40

For the second interface, the law of refraction is


sin r1 = n2 sin i2 = 1.50 sin 25.40 = 0.6434 n1 1.00

and the angle of refraction for the second interface is

r1 = sin−1 (0.6434)r1 = 40.00

Note that the final ray makes the same angle as the initial ray, and they are,therefore, parallel.


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31.5 Total Internal ReflectionAs shown in figure 31.1, when an incident ray falls on a boundary of a transparentmaterial, part of the ray is reflected back into the first medium, while part of theray is refracted through the second medium. If the incident ray is in the densemedium, then the refracted ray is bent away from the normal, as shown in figure31.6(a). If the angle of incidence is increased, the angle of refraction is alsoincreased, as shown in figure 31.6(b). If the angle of incidence is increased still

Figure 31.6 Total internal reflection.

further, the point is reached where the angle of refraction becomes 900, as shown infigure 31.6(c). The angle of incidence that causes the refracted ray to bend through900 is called the critical angle of incidence. When the incident angle becomesgreater than the critical angle, no refraction occurs. That is, no light enters thesecond medium at all; all the light is reflected. This condition is called total internalreflection, because refraction has been eliminated entirely. We obtain the conditionfor total internal reflection by finding the critical angle of incidence that allows theangle of refraction to be 900. This is done by applying Snell’s law of refraction:

n1 sin ic = n2 sin 900

orsin ic = n2 sin 900 = n2 (31.17)

n1 n1

Equation 31.17 gives the condition for total internal reflection.Total internal reflection can only occur when light travels from a denser

medium to a rarer medium, because it is only then that the refracted ray is bent


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away from the normal. For a light ray going in the opposite direction, the refractedray bends toward the normal and the angle of refraction could never become 900.

Example 31.4

Critical angle for a glass-air interface. What is the critical angle for a ray of lightthat goes from glass to air? Take nglass = 1.50.

Solution

The critical angle, found from equation 31.17, is

sin ic = n2 = nair = 1.00 = 0.667 n1 nglass 1.50

ic = sin−1 (0.667)ic = 41.80


Let us now take an ordinary piece of glass, whose index of refraction is 1.50,and cut it into a triangle with angles of 45.00, 90.00, and 45.00, as shown in figure31.7 An incident ray perpendicular to the first face goes directly into the glass

Figure 31.7 Total internal reflection and the prism.

without any refraction, because it is along the normal of the first face. This raymakes an angle of incidence of 45.00 with the normal as it hits the second face. But45.00 is greater than the critical angle of 41.80 just found in example 31.4. Hence,none of this light crosses the interface into the air, but instead it is totally reflectedat this second face. This reflected ray, being normal to the third face, completely


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passes into the air at the third face without any refraction there. This ordinary pieceof glass, cut into the shape of a triangle, with an angle greater than the critical angle,is called a prism. One of its functions is to completely reflect light. It is even betterthan a plane mirror for reflection because in a plane mirror some light energy isabsorbed by the silver of the mirror, whereas the prism, acting as a device for totalinternal reflection, reflects everything. No energy is ever transmitted into the othermedium at the interface. Prisms are used in binoculars, spectrometers, andreflecting telescopes.

Example 31.5

Critical angle for a water-air interface. What is the critical angle between water andair? The index of refraction for water is nw = 1.33.

Solution

The critical angle, found from equation 31.17, is

sin ic = n2 = nair = 1.00 n1 nw 1.33

ic = 48.80

The critical angle for a water-air interface is thus greater than the critical angle fora glass-air interface.


A further example of internal reflection can be found in the area of fiberoptics. An optical fiber consists of a single flexible glass rod, or an array of them, ofhigh refractive index. Light entering the glass undergoes total internal reflectionfrom the walls of the glass fiber and the light travels down the length of the fiberwith little or no absorption. Due to their flexibility, the fibers can be curved intovarious shapes. But as long as the reflection angle remains greater than the criticalangle, the light travels along the length of the fiber.

31.6 DispersionIn all the previous discussions of refraction we assumed that the incident light wasmonochromatic, that is, it contained light of only one wavelength. If white light, amixture of all colors and hence numerous wavelengths, impinges on a transparentsurface, we obtain the refraction pattern shown in figure 31.8. The white light is


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Figure 31.8 Dispersion of light.

refracted such that each color, or wavelength, is refracted by a different angle.Hence, the white light is dispersed into its constituent wavelengths. The separationof white light into its component colors is called dispersion. The band of colors isknown as a spectrum. The spectrum of visible light contains the colors red, orange,yellow, green, blue, and violet.

Violet light, which has the shortest wavelength, is bent the most, whereas redlight with the longest wavelength, is bent the least. Dispersion occurs because theindex of refraction is not strictly a constant for a particular material, but rathervaries slightly with wavelength. For example, the index of refraction of crown glassfor red light (700.0 nm) is about 1.51, whereas for violet light (400.0 nm) it is about1.53, which is only a slight difference, yet significant enough to cause thephenomenon of dispersion.

Example 31.6

Refraction for various wavelengths. White light makes an angle of incidence of 30.00

as it strikes a piece of glass. Find the angle of refraction for (a) red light (700.0 nm)if n = 1.51 for red light and (b) for violet light (400.0 nm) if n = 1.53.

Solution

a. The angle of refraction for red light, found from Snell’s law, is

n1 sin i = n2 sin rr

sin rr = n1 sin i = 1.00 sin 30.00 = 0.331 n2 1.51

rr = sin−1 (0.331) rr = 19.30

b. For violet light, the angle of refraction is

n1 sin i = n2 sin rv


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sin rv = n1 sin i = 1.00 sin 30.00 = 0.3267 n2 1.53

rv = sin−1 (0.3267)rv = 19.10

Notice that the total angular separation between the red and violet light is only 0.2degrees for this glass.


Dispersion is most pronounced if a prism is used, as shown in figure 31.9.Because of the angle between the two faces of the prism, the deviation D of the finalrefracted ray is greater.

Figure 31.9 Dispersion by a prism.

Because the index of refraction is a function of the incident wavelength, whendefining the index of refraction of a medium, we need to specify the particularwavelength used. When the wavelength is not specified in this book it is assumedthat the wavelength is λ = 589.2 nm, corresponding to the yellow D line of thesodium spectrum.

31.7 Thin LensesAn optical lens is a piece of transparent material, such as glass or plastic. Because ofits shape, however, light passing through it either converges or diverges to itsprincipal axes. The entire effect of the lens is due to its shape and the index ofrefraction of the lens.

One such shape of a lens is formed from a piece of plane glass by shaping itinto the form of two spherical surfaces with radii of curvature R1 and R2, as shownin figure 31.10, where C1 and C2 are the centers of the two spherical surfaces. Thisshaped piece of glass is called a double convex lens, because the shape of its surfacesare convex.


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Figure 31.10 Formation of a lens.

The line going through the center of the lens is called the principal axis and isshown in figure 31.11. Consider a ray of light AB, parallel and close to the principal

Figure 31.11 Refraction by a lens.

axis and impinging on the first surface at the point B. The normal N1 to the surfaceat B is a continuation of the radius of curvature of the first surface, which emanatesfrom the center of curvature C1 of the first surface. The incident ray AB makes anangle of incidence i1 with the normal N1. Because the ray is going from the lessdense medium, air, to the more dense medium, glass, the refracted ray is bentthrough an angle r1 toward the normal, as shown. This refracted ray impinges onthe second surface at the point D. The normal at D is the continuation of the radiusof curvature of the second surface and is designated N2. Ray BD makes an angle ofincidence i2 with the normal N2. Since the ray now goes from glass to air, it is goingfrom a more dense medium to a less dense medium. The refracted ray is, therefore,bent away from the normal N2 through an angle r2 as ray DE. The net effect of the


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two refractions is to take a ray of light, which is parallel to the principal axis, andbend it such that it crosses the principal axis. The point where this ray crosses theprincipal axis is called the principal focus, and is designated by the letter F. Thedistance from the center of the lens to the principal focus is called the focal length fof the lens. It is important to note that for a particular material, it is the shape ofthe lens that determines the normals and hence the refracted rays and thus thefocal length of the particular lens.

Since we have already seen that the converging of a parallel ray to theprincipal axis is caused by the refraction at the two surfaces of the lens, it will notbe necessary to go into all this detail every time we consider a lens. Instead, we willsay that a ray of light, parallel to the principal axis, is bent in the middle of thelens. It then converges to the principal focus, as shown in figure 31.12.

Figure 31.12 Bending of a ray of light by a convex lens.

All rays parallel and close to the principal axis converge to the principalfocus, as seen in figure 31.13. Such a lens is called a converging lens. If theincident rays are not close to the principal axis, the converging rays do not all

Figure 31.13 A converging lens.

intersect at the same point, a result known as spherical aberration. We will assumein this book that all rays are sufficiently close to the principal axis so that sphericalaberration can be ignored. If parallel light comes in from the right of the lens itconverges at the same distance to the left of the lens. Note also that, by theprinciple of reversibility, light emanating from the principal focus comes out parallel


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to the principal axis. Figure 31.14 shows some examples of converging lenses. Theactual shapes are determined by the radii of curvature, but an easy way to identifysuch a lens is to note that a converging lens is always thicker at the center of the lens

Figure 31.14 Examples of converging lenses.

than it is at the rim. If any of the thin lenses in figure 31.14 are reversed, they stillproduce the same effect.

A diverging lens is a lens that takes a bundle of parallel light rays anddiverges them away from the principal axes, as shown in figure 31.15. If the

Figure 31.15 A diverging lens.

diverging rays are produced backward, they all intersect at the same point F, calledthe principal focus, or focal point, of the diverging lens. The diverging rays appear tocome from the principal focus. The distance from the center of the lens to the focalpoint is called the focal length f of the diverging lens. Some examples of diverginglenses are shown in figure 31.16. One characteristic of all diverging lenses is thatthey are thinner at the center of the lens than they are at the rim.

Figure 31.16 Examples of diverging lenses.


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31.8 Ray Tracing and the Standard RaysTo determine the location of an image formed by a lens, we use the technique of raytracing. An object OP, represented by a brown arrow, is placed a distance p in frontof a convex lens in figure 31.17. A ray, from the bottom of the object at P, travels

Figure 31.17 Ray diagram for a convex lens.

along the principal axis. This ray goes through the lens undeviated. Hence thebottom of the image must lie somewhere on the principal axis of the lens. Todetermine the location of the tip of the image, three rays are drawn from the tip ofthe object. These three rays are called the standard rays and are shown in figure31.17.

The first ray (1), in red, is drawn parallel to the principal axis and afterrefraction it passes through the principal focus. A second ray (2), in green, is drawnfrom the top of the object and passes through the first focal point. This ray isrefracted through the lens so that it comes out parallel to the principal axis. (Recallthat by the principle of reversibility, light emanating from the principal focus comesout parallel to the principal axis.) The third of the standard rays (3), in blue, isdrawn from the tip of the arrow and goes through the center of the lens. Therefraction experienced at the first surface is exactly compensated by the refractionat the second surface, and the ray passes through the lens undeviated. These threerays intersect at the point I and form the real, inverted image IQ, in yellow. If ascreen were placed at this point a sharp image would be found there. The distancefrom the object to the lens is called the object distance p, whereas the distance fromthe lens to the image is called the image distance q. Thus, we can find any image bydrawing a ray diagram.

A ray diagram for a diverging lens is shown in figure 31.18. An object OP isplaced a distance p in front of a diverging lens. A ray (1), in red, is drawn from theobject parallel to the principal axis. The diverging lens causes the ray to bend awayfrom the principal axis, as if the ray had come from the principal focus F. A secondray (2), in green, is drawn, which would go to the second focal point if there were no


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Figure 31.18 Ray diagram for a concave lens.

lens present. This ray is caused to bend by the diverging lens so that it comes outparallel to the principal axis. A third ray (3), in blue, passes through the center ofthe lens and is not bent. These three rays do not intersect anywhere to the right ofthe lens and hence cannot form a real image there. However, if we follow each of thethree rays backward, they appear to intersect at the point I, and the image islocated at that point. The distance from the lens to the image, the image distance, isdenoted by the letter q in the diagram. We can find an image of any object for anylens by drawing these standard rays.

31.9 The Lens EquationA mathematical relation between the object distance p, the image distance q, andthe focal length f, can be obtained from the geometry of figure 31.19.

In triangle OPC,tan θ1 = OP = ho

PC p whereas in triangle QIC,

tan θ1 = QI = hi

QC q

Equating the tangent of θ1 from each equation gives

ho = hi (31.18) p qRearranging terms, this becomes


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Figure 31.19 Determining the lens equation.

hi = q (31.19) ho pWe will return to this equation shortly.

In triangle ACF,tan θ2 = AC = ho

CF fwhereas in triangle IQF,

tan θ2 = QI = hi QF q − f

Equating the tangent of θ2 from each equation, gives

hi = ho q − f fRearranging terms, this becomes

hi = q − f (31.20) ho f

Setting equation 31.20 equal to equation 31.19, gives

q − f = q f por

q − 1= q f pDividing each term by q, gives

1 − 1 = 1 f q pSolving for 1/f, we obtain


31-22

1 = 1 + 1 (31.21) f p q

Equation 31.21 is the lens equation and gives the relation between the objectdistance p, the image distance q, and the focal length f of the lens. Notice that it isthe same formula as the mirror equation found in chapter 30.

The linear magnification M of a lens system is defined as the ratio of theheight of the image to the height of the object, that is,

M = height of image = hi (31.22) height of object ho

Using equation 31.19, we can also write this as

M = hi = − q (31.23) ho p

The magnification tells how much larger the image is than the object. When weknow M, we find the height of the image hi from equation 31.23, as

hi = Mho (31.24)

The minus sign in equation 31.23 is placed there by convention. For a single lens,when the magnification is negative the image is real and inverted, when it ispositive the image is virtual and erect. Notice that the lens equation, themagnification equation, and the height of the image equation, are the sameequations that were obtained for the spherical mirror in chapter 30.

Example 31.7

A converging lens. An object 5.00 cm high is placed 35.0 cm in front of a converginglens of 10.0 cm focal length. (a) Where is the image located? (b) What is themagnification? (c) What is the size of the image?

Solution

a. The image distance, found from the lens equation, equation 31.21, is

1 = 1 + 1 f p q

1 = 1 − 1 = 1 − 1 = 0.0714 cm−1

q f p 10.0 cm 35.0 cmq = 14.0 cm


31-23

b. The magnification, found from equation 31.23, is

M = − q = −14.0 cm = −0.400 p 35.0 cm

Since M is negative, the image is real and inverted.

c. The size of the image, found from equation 31.24, is

hi = Mho = −(0.400)(5.00 cm) = −2.00 cm

The image is thus smaller than the object. The minus sign on hi simply means thatthe distance hi is in the negative y direction, and hence, the image is inverted.


Example 31.8

A diverging lens. An object 5.00 cm high is placed 35.0 cm in front of a diverginglens of −10.0 cm focal length. (a) Where is the image located? (b) What is themagnification? (c) What is the size of the image?

Solution

a. The image distance, found from the lens equation, equation 31.21, is

1 = 1 + 1 f p q

1 = 1 − 1 = 1 − 1 q f p −10.0 cm 35.0 cm

= −0.100 cm−1 − 0.0285 cm−1 = −0.1286 cm−1

q = −7.78 cm

Since the image distance q is negative, the image is virtual and is located 7.78 cm infront of the lens.


= +0.222M = −qp = − −7.78 cm

35.0 cm


31-24


Because M is positive the image is virtual and erect.

c. The size of the image, found from equation 31.24, is

hi = Mho = (0.222)(5.00 cm) = 1.11 cm


31.10 Some Special Cases for the Convex LensCase 1: The Object Is Located at InfinityWhen the object is very far away from the lens, we can assume that it is at infinity.Setting p = ∞ in the lens equation, gives, for the location of the image,

1 = 1 − 1 q f p

= 1 − 1 f ∞

But 1/∞ = 0, and the equation reduces to

q = f

That is, when the object is located at infinity, the image is located at the principalfocus. The magnification becomes

M = − q = − q = 0 p ∞

which shows that the image is reduced to a point. This case, which was shown infigure 31.13, is used as a simple technique to determine the focal length of aconverging lens. A converging lens is held in front of a card and the distancebetween the card and the lens is varied until a sharp image is obtained on the cardof a very distant object ( p equal to infinity). The distance between the lens and thecard is the image distance q. But as just seen, if the object is at infinity, then q = f.Hence, the distance from the lens to the card is the focal length of the converginglens.

Case 2: The Object Lies between Infinity and 2f, That Is, 2f < p <∞∞∞∞

This is a rather general case and the ray diagram is shown as the continuous lines


31-25


in figure 31.20. The actual location of the image is found from the lens equationwhen p and f are specified.

Figure 31.20 The object moves closer to the lens.

A very interesting result occurs when the object is moved in toward the lens.The ray diagram for this case is shown in dashed lines in figure 31.20. Notice that asthe object moves toward the lens the image moves away from the lens and gets bigger.However, the height of the image hi never gets greater than the height of the objectho, and thus the magnification is always less than 1. (This will be proved in case 3.)

Case 3: The Object Is Located at a Distance of Twice the FocalLength from the Lens, That Is, p = 2fFor this case the image, found from the lens equation, is

1 = 1 + 1 = 1 + 1 f p q 2f qor

1 = 1 − 1 = 1 q f 2f 2f and hence,

q = 2f

That is, when the object is placed a distance of 2f to the left of the lens, the image islocated at the same distance of 2f to the right side of the lens. The magnification forthis case becomes

M = hi = − q = − 2f = −1 ho p 2f and

hi = −ho


31-26

Hence, when p = q = 2f, the height of the image is the same size as the height of theobject. For any value of p greater than 2f the magnification is always less than 1, andhence the height of the image is less than the height of the object. For any value of pless than 2f, the magnification is greater than 1, and hence the height of the image isalways greater than the height of the object.

Case 4: The Object Is Located between the Focal Length andTwice the Focal Length, That Is, f < p < 2fIt is within this region that the lens acts as a magnifier, because if p is less than 2f,the magnification is greater than 1. The ray diagram is shown in figure 31.21. It isobvious from the diagram that q is greater than p, and hence, the magnification, M

Figure 31.21 Ray diagram for object located between the focal length and twice thefocal length.

= q/p, is greater than 1, and the image is enlarged. This is also seen in the diagram,that is, the height of the image is greater than the height of the object.

Case 5: The Object Is Located at the Principal Focus of the Lens,That is, p = fWhen the object is placed at the principal focus, p = f, and the lens equation givesfor the location of the image,

1 = 1 − 1 = 1 − 1 = 0 q f p f f

The only way for 1/q to be equal to zero, is for q to equal infinity, that is,

q = ∞


31-27

The ray diagram for such a case is shown in figure 31.22. The standard ray(2) cannot be drawn, but rays (1) and (3) can, and as we see from the diagram, theybecome parallel on the right-hand side of the lens and hence never intersect to forma real image there at any finite distance from the lens. The parallel rays onlyintersect at infinity and it is there where the image may be found. Note that if rays(1) and (3) were dashed backward, they would still be parallel, and hence, could notform a virtual image either.

Figure 31.22 Ray diagram for an object at the principle focus.

Case 6: The Object Is Placed within the Principal Focus, That Is,p < fA ray diagram for this case is shown in figure 31.23. Standard ray (1) is drawn

Figure 31.23 Ray diagram for an object placed within the principle focus.


31-28

parallel to the principal axis and is refracted in the lens such that it passes throughthe principal focus F, as shown. Of all the infinite rays that emanate from the tip ofthe arrow, standard ray (2) is a ray that is aligned with the dashed straight linefrom the principal focus to the tip of the arrow of the object. Therefore, ray (2)appears to come from the first principal focus, and hence, is refracted in the lenssuch that it comes out parallel to the principal axis. The third standard ray (3) isdrawn from the tip of the object and goes through the center of the lens undeviated,as shown. The three standard rays do not converge anywhere on the right-hand sideof the lens to form a real image there. However, if these standard rays are producedbackward they intersect on the left-hand side of the lens and form a virtual imagethere. Thus a person looking through the lens from the right would see theenlarged, erect, virtual image of the object on the left-hand side of the lens. When aconvex lens is used with the object located within the principal focus, it is called asimple magnifying glass. In this case, the image distance q is always negative.

Example 31.9

A simple magnifying glass. An object 5.00 cm high is placed 3.00 cm in front of aconvex lens of 5.00-cm focal length. (a) Where is the image located? (b) What is themagnification? (c) How high is the image?

Solution

a. The image distance, found from the lens equation, is

1 = 1 − 1 = 1 − 1 = −0.1333 cm−1

q f p 5.00 cm 3.00 cmand

q = −7.50 cmNote that q is a negative quantity.


= +2.50M = −qp = − −7.50 cm

3.00 cm

The magnification is thus positive and indicates that the image is virtual and erect.

c. The image height, found from equation 31.24, is

hi = Mho

= (2.50)(5.00 cm)= 12.5 cm


31-29

Thus the lens used in this way gives an enlarged (magnified 2.5 times), erect,virtual image of the object. The fact that q is negative is an indication that theimage is virtual.


In summary, for a converging lens, with the object distance p positive, if theimage distance q is positive the image is real and located on the other side of thelens. If the image distance q is negative, the image is on the same side of the lens asthe object and is a virtual image.

31.11 Combinations of Lenses

Most optical systems consist of a combination of two or more lenses, prisms, ormirrors. Figure 31.24 shows a combination of two convex lenses separated by a

Figure 31.24 Combination of two convex lenses.


31-30


distance d. The first lens has a focal length f1 and the second lens has the focallength f2, as illustrated. An object of height ho1 is placed a distance p1 in front of thefirst lens. Let us find the location and size of the resulting image of the lenscombination. We begin the analysis by considering the first lens only, that is, wetemporarily ignore the existence of the second lens. We find the image distance q1 ofthe first lens by drawing the standard rays of a ray diagram, as in figure 31.24. Thisimage of the first lens now acts as the object of the second lens. We now completelyignore the existence of the first lens and consider the second lens as though anobject has been placed a distance p2 in front of it. We again find the image from thissecond lens by drawing the standard rays of a ray diagram, as shown in figure31.24. Thus, the image distance q2 of the second lens and the image height hi2 areeasily found.

In addition to using the ray diagram, we can also find the final image fromthe lens equation. The image distance of the first lens is found from

1 = 1 − 1 (31.25) q1 f1 p1

whereas the image distance of the second lens is found from

1 = 1 − 1 (31.26) q2 f2 p2

However, as we can see from the diagram, the object distance p2 is

p2 = d − q1 (31.27)

In most of the cases that we consider in this book, the image distance q1 turns out tobe less than the separation distance d, and the term (d − q1) is positive, giving apositive object distance p2. There are some cases where the image distance q1 isgreater than the separation distance d. In these cases, the term (d − q1) is anegative quantity and hence, the object distance p2 is also a negative quantity. Anobject having a negative object distance is called a virtual object. For a virtualobject, the rays strike the lens before the object is formed. We will see an example ofthis later.

The magnification of the lens combination is defined as the ratio of the heightof the final image to the height of the initial object. That is,

M = height of image 2 = hi2 (31.28) height of object 1 ho1

But the magnification of the first lens is


31-31

M1 = height of image 1 = hi1 (31.29) height of object 1 ho1

whereas the magnification of the second lens is

M2 = height of image 2 = hi2 (31.30) height of object 2 ho2

Thus, from equation 31.30,hi2 = ho2M2 (31.31)

and from equation 31.29,ho1 = hi1 (31.32)

M1

Substituting equations 31.31 and 31.32 back into equation 31.28, and noting thatthe height of image 1 hi1 is equal to the height of object 2 ho2, we get

M = hi2 = ho2M2 = ho2M2 ho1 hi1/M1 ho2/M1

orM = M1M2 (31.33)

Equation 31.33 says that the magnification of a combined optical system is equal tothe product of the magnification of each individual lens. The height of the finalimage, obtained from equation 31.28, is

hi2 = Mho1 (31.34)

The location and nature of the image of a particular optical system dependson the focal lengths f1 and f2, the location of the first object p1, and the separation dbetween the two lenses. We will consider different optical systems in section 31.13.

Example 31.10

A combination of lenses. An object 5.00 cm high is placed 15.0 cm in front of aconvex lens of 10.0-cm focal length. A second convex lens, also of 10.0-cm focallength, is placed 48.0 cm behind the first lens. Find (a) the image of the lenscombination, (b) the magnification of the system, and (c) the height of the finalimage.

Solution

a. The image of the first lens, found from equation 31.25, is


31-32

1 = 1 − 1 = 1 − 1 = 0.0333 cm−1 q1 f1 p1 10.0 cm 15.0 cm

q1 = 30.0 cm

The object distance for the second lens, found from equation 31.27, is

p2 = d − q1 = 48.0 cm − 30.0 cm = 18.0 cm

Hence, the final image, found from equation 31.26, is

1 = 1 − 1 = 1 − 1 = 0.0444 cm−1 q2 f2 p2 10.0 cm 18.0 cm

q2 = 22.5 cm

Thus, the final image of the lens combination is found 22.5 cm behind the secondlens, or 70.5 cm behind the first lens.

b. The magnification of each lens is found as

M1 = − q1 = − 30.0 cm = −2.00 p1 15.0 cm

M2 = − q2 = − 22.5 cm = −1.25 p2 18.0 cm

The final magnification of the system is

M = M1M2 = (−2.00)(−1.25) = 2.50

Note that M1 is negative indicating that the first lens inverts the first image. Thefact that M2 is also negative means that lens 2 inverted the second image fromupside down to erect. The total magnification M is positive indicating that the finalimage is right side up. In dealing with combinations of lenses we must be careful ofthe convention adopted for the magnification of a single lens. The conventionadopted is that if M is negative, the image is real and inverted, whereas if M ispositive, the image is erect and virtual. In this case of the combined lens, M1 isnegative and image 1 is real and inverted, as expected. Also M2 is negative andimage 2 is real and inverted from its original orientation when in front of lens 2,also as expected. The inversion of the inverted image, gives a final image that iserect. Notice however, that the final magnification is positive because it is theproduct of two negative numbers. But the final image is real not virtual. Hence, theconvention that a positive magnification indicates that the image must be virtualcannot be used for a system of multiple lenses.


31-33

c. The final height of the image of the system, found from equation 31.34, is

hi2 = Mho1 = (2.50)(5.00 cm)= 12.5 cm


31.12 Thin Lenses in ContactAn interesting case results from the lens combination of section 31.11 when thedistance d between the two lenses is made equal to zero, thereby placing the twolenses in contact with each other. The lens equation applied to the second lens is

1 = 1 + 1 (31.35)f2 p2 q2

Setting p2 = d − q1 from equation 31.27 into equation 31.35, yields

1 = 1 + 1 (31.36) f2 d − q1 q2

Setting the separation distance d equal to zero in equation 31.36 yields

1 = 1 + 1 (31.37) f2 − q1 q2

But the lens equation applied to the first lens is given by

1 = 1 − 1 q1 f1 p1

and 1 = − 1 + 1 (31.38)

− q1 f1 p1

Substituting equation 31.38 into equation 31.37, yields

1 = − 1 + 1 + 1 f2 f1 p1 q2

and


31-34


1 + 1 = 1 + 1 (31.39) f1 f2 p1 q2

Let us examine equation 31.39 and see what it is telling us. Here p1 is the objectdistance of the first lens, but it is also the object distance of the lens combination,and we will designate it as pc. Although q2 is the image distance of the second lens,it is also the image distance of the lens combination, and we can call it qc. Hence, wecan rewrite equation 31.39 as

1 + 1 = 1 + 1 (31.40) f1 f2 pc qc

The right-hand side of equation 31.40 looks like part of the lens equation, so it isonly natural to define the focal length for the two lenses in contact as fc, and thelens equation for the combination should then be

1 = 1 + 1 (31.41) fc pc qc

Combining equations 31.40 and 31.41, gives

1 = 1 + 1 (31.42) fc f1 f2

The focal length of the combination is thus obtained from equation 31.42. Hence,whenever two lenses are in contact, the reciprocal of the focal length of thecombination is equal to the sum of the reciprocals of the focal lengths of theindividual lenses.

Example 31.11

The focal length of two convex lenses in contact. A convex lens of 10.0-cm focal lengthis placed in contact with a convex lens of 15.0-cm focal length. What is the focallength of the combination?

Solution

The focal length of the combination, found from equation 31.42, is

1 = 1 + 1 = 1 + 1 = 0.1666 cm−1 fc f1 f2 10.0 cm 15.0 cmand

fc = 6.00 cm


31-35


Example 31.12

The focal length of a convex and a concave lens in contact. A 10.0-cm convex lens isplaced in contact with a 15.0-cm concave lens. What is the focal length of thecombination?

Solution

The focal length of the combination is found from equation 31.42, but since thesecond lens is concave its focal length is negative, that is, f2 = −15.0 cm. Therefore

1 = 1 + 1 = 1 − 1 = 0.0333 cm−1 fc f1 f2 10.0 cm 15.0 cmand

fc = 30.0 cm


Equation 31.42, although derived for only two lenses, is completely generaland thus, for any number of lens placed in contact the focal length of the combinationis

1 = 1 + 1 + 1 + 1 + … (31.43) fc f1 f2 f3 f4

When dealing with many lenses in contact it is convenient to define thedioptric power of a lens as the reciprocal of the focal length in meters. Thus

P = 1 (31.44) f

The reciprocal of the focal length in meters is called a diopter.

Example 31.13

The power of a lens. What is the power of a 15.0-cm focal length lens?

Solution


31-36



The focal length of the lens is f = 15.0 cm = 0.150 m. The power of the lens, foundfrom equation 31.44, is

P = 1 = 1 = 6.67 diopters f 0.150 m


The advantage of expressing a lens in terms of its dioptric power is evidentwhen the lenses are in contact. For example, if there are four lenses of focal lengthsf1, f2, f3, and f4, then the power of each lens is

P1 = 1 f1

P2 = 1 f2

P3 = 1 f3

P4 = 1 f4

The power of these lenses in combination, determined from equation 31.43, is

1 = 1 + 1 + 1 + 1 fc f1 f2 f3 f4

and substituting the values of 1/f, we get

1 = P1 + P2 + P3 + P4

fc

The power of the combination is thus

Pc = P1 + P2 + P3 + P4 (31.45)

Expressing a lens in terms of dioptric power is convenient for an optician,who places many different lenses in contact. When a patient is given an eye exam,the optician places different lenses in front of the patient’s eyes and asks him toread a chart. When the patient finds the sharpest image, the optician immediatelygets the power of the combination of lenses used, Pc, from equation 31.45. Thereciprocal of Pc is the focal length of the eyeglasses needed for the patient and a lensof this focal length can be ground using the lensmaker’s formula, which we nowdiscuss.


31-37


31.13 The Lensmaker’s FormulaWe showed in figure 31.10 that a spherical lens was made from a piece of planeglass by shaping it into the form of two spherical surfaces with radii of curvature R1

and R2. We then showed in figure 31.11 that a ray of light parallel to the principalaxis of this lens was refracted and then crossed the principal axis at the focal pointF. Then in figure 31.13 we showed that all rays parallel and close to the principalaxis converge to the principal focus, and we defined the distance from the center ofthe lens to the focal point as the focal length f of the lens. We then showed in section31.10 that the focal length of any lens could be easily determined experimentally byplacing a converging lens in front of a card and the distance between the card andthe lens is varied until a sharp image is obtained on the card of a very distant object( p equal to infinity). The distance between the lens and the card is the imagedistance q. But if the object is at infinity, then q = f. Hence, the distance from thelens to the card is the focal length of the converging lens. In this way we coulddetermine the focal length of any lens experimentally. But how do we know how togrind the lens into the spherical shape that will gives us this focal length?

Recall from chapter 30 that the focal length of a spherical mirror was foundby equation 30.14 to be f = R/2, where R was the radius of curvature of the mirror.That is, if you want a particular focal length f, just grind the mirror to the radius R= 2f. In this way you can make a mirror of any focal length. In a similar way, wewould now like to determine the focal length of the lens in terms of its radii ofcurvature and the index of refraction of the lens material. The equation that relatesthe focal length, index of refraction, and radii of curvature of a lens is called thelensmaker’s formula. To derive this relation we place an object at the point A, atthe object distance p, in a medium of index of refraction n1, in front of a thick lens asseen in figure 31.25. We consider the ray of light that leaves the object at A and

p2

n2 n1n1

p1q1

q2

= p2 q1t +

A

B

I

E

F

Figure 31.25 Refraction of a ray of light through a thick lens.


31-38

strikes the spherical surface at the point B. The incident ray AB, upon entering thesecond medium of index of refraction n2, is bent toward the principal axis and thisrefracted ray crosses the principal axis at the point I, and at this point representsthe image I of the object at A. The distance from the front of the lens to the point I isthe image distance q1. The ray then continues through the second medium until itarrives at the point E where it is again refracted toward the principal axis andintersects the principal axis at the point F, which will be the final image of the thicklens. The distance from the point I to the right surface of the lens is the objectdistance for the second spherical surface and is designated as p2. The distance formthe right surface to the final image F is the image distance q2 of the second sphericalsurface. Let us now look at what happens at each of the spherical surfaces of thethick lens.

The incident ray leaves the object at A and strikes the spherical surface atthe point B. The normal at the point B is drawn as the extension of the radius R1 ofthe spherical surface whose center of curvature is located at the point C1, figure31.26. The incident ray AB makes an angle of incidence i1 with the normal N. Uponentering the second medium of index of refraction n2, the light ray is bent towardthe normal making the angle of refraction r1 as shown in the figure. This refractedray crosses the principal axis at the point I, and represents the image I of the objectat A. The law of refraction applied to the interface at point B gives

n1 sin i1 = n2 sin r1 (31.46)

n2n1

p1 q1

ACD

Bδ

θα γβ

1i1r

I

N

1R

1R

Figure 31.26 Refraction of a ray of light at the first spherical surface.

We consider only incident rays that are close to the principal axis. Such rays arecalled paraxial rays. For paraxial rays, the angle α and hence i1 and r1 will all besmall angles2 and hence the sine of these angles will be equal to the angle itself.That is, we let sin i1 = i1 and sin r1 = r1, hence the law of refraction becomes


31-39

2The angles in our diagrams are greatly exaggerated to show the details of the refraction.

n1 i1 = n2 r1 (31.47)

Recall from geometry that the exterior angle of a triangle is equal to the sum of thetwo interior angles. Applying this result to triangle ABC we see that

i1 = α + θ (31.48)and for triangle BIC,

θ = r1 + γSolving for r1 gives

r1 = θ − γ (31.49)

Substituting equations 31.48 and 31.49 into equation 31.47 gives

n1 (α + θ) = n2 (θ − γ) (31.50)

Now recall that the arc s of a circle is given by s = rθ. Applying this relation to arcBD of the lens we get

BD = R1θand hence

θ = BD (31.51) R1

Because the ray AB is a paraxial ray, the arc BD is approximately equal to q1γ, thatis

BD ≈ q1γand hence

γ = BD (31.52) q1

Also arc BD is approximately equal to p1α, that is

BD ≈ p1αand hence

α = BD (31.53) p1

Replacing equations 31.51, 31.52, and 31.53 into equation 31.50 gives

n1 (BD + BD) = n2 (BD − BD) p1 R1 R1 q1

Canceling out the term BD on both sides of the equation gives


31-40

n1 + n1 = n2 − n2

p1 R1 R1 q1

Rearranging terms yields n1 + n2 = n2 − n1

p1 q1 R1 R1

and n1 + n2 = n2 − n1 For first spherical surface (31.54)

p1 q1 R1

Let us now consider what happens at the second spherical surface. The ray oflight from the first surface that converged to the point I as the image of the firstsurface now becomes the object for the second spherical surface, figure 31.27. The

p2 q2

n2n1

FG

Eδ

θ αγ βI

N

i2r2

R2

R2C2

Figure 31.27 Refraction of a ray of light at the second spherical surface.

distance from the point I to the right surface of the lens is the object distance for thesecond spherical surface and is designated as p2. The distance from the right surfaceto the final image F is the image distance q2 of the second spherical surface. Usingthe same analysis on the right spherical surface as on the left spherical surface weobtain the equivalent relation of equation 31.54 as

n2 + n1 = n1 − n2 For second spherical surface (31.55) p2 q2 R2

The combined effects of both refractions are found by adding equations 31.54 and31.55 as

n1 + n2 + n2 + n1 = (n2 − n1) + (n1 − n2) p1 q1 p2 q2 R1 R2

n1 + n1 + n2 + n2 = (n2 − n1) − (n2 − n1) p1 q2 p2 q1 R1 R2

(31.56)n1p1

+ n1q2 + n2

1p2

+ 1q1 = (n2 − n1 ) 1

R1− 1

R2


31-41

Let us now analyze the third term in equation 31.56.

(31.57)n21p2 + 1

q1 = n2q1 + p2p2q1 = n2t

p2q1

where we have used the fact that the sum q1 + p2 = t, the thickness of the lens asseen in figure 31.25. If we make the lens a thin lens then in the limit we canconsider the thickness t of the lens to be small compared to the object and imagedistances and hence we let t approach zero. That is, as t → 0, the term in equation31.57 approaches zero and can be dropped from equation 31.56. Thus equation31.56 becomes

n1p1

+ n1q2 = (n2 − n1 ) 1

R1− 1

R2or

(31.58)1p1

+ 1q2 = n2 − n1

n11

R1− 1

R2

The object distance p1 for the first spherical surface is also the object distance of theentire lens p and so we let p1 = p, and q2 the image distance of the second sphericalsurface is the image distance q of the entire lens and so we now let q2 = q. Thereforeequation 31.58 becomes

(31.59)1p + 1

q = n2 − n1n1

1R1

− 1R2

But we showed in equation 31.21, the lens equation, that

(31.21)1p + 1

q = 1f

Setting equation 31.21 equal to equation 31.59 gives

(31.60)1f = n2 − n1

n11

R1− 1

R2

Equation 31.60 is called the lensmaker’s equation because it allows the lens maker tomake a lens of a particular focal length f by the appropriate choice of radii ofcurvature R1 and R2 of the spherical lens and the index of refraction n1 of the mediumin which the lens will be used and n2 the index of refraction of the lens itself. If thelens is to be used in air, then n1 = 1 and n2 = n, the index of refraction of the lens,and equation 31.60 becomes

(31.61)1f = (n − 1) 1

R1− 1

R2

Equation 31.61 is the lensmaker’s formula for a lens used in air, n is the index ofrefraction of the glass, R1 is the radius of curvature of the first surface of the lens, R2

is the radius of curvature of the second surface, and f is the focal length of the lens.


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By convention, if the surface is convex to the incident ray, R is consideredpositive. If the surface is concave to the incident ray, R is considered negative. It isassumed that the ray of light is incident from the left. Thus, for a double convexlens, R1 would be positive and R2 would be negative. Equation 31.61 says that inorder to make a lens of a particular focal length f for a medium of index of refractionn, the medium must be ground to the radii of curvature R1 and R2 that satisfiesequation 31.61. Note that in the derivation of equation 31.61, we assumed that thethickness t of the glass lens is negligible compared to the distance to the principalfocus and to any object or image distance concerned. Such a lens of negligiblethickness is called a thin lens. We have only considered thin lenses in this book.

The focal length of any converging or diverging lens is found from thelensmaker’s formula, equation 31.61. However, it is important to remember theconvention that R is positive for an incident ray that impinges on a convex surfaceand negative for one that impinges on a concave surface. A plane surface is a surfacewhose radius of curvature is infinite, and hence, 1/R = 0.

Example 31.14

Finding the focal length of a converging lens. A double convex lens has radii ofcurvature of 25.0 cm. The index of refraction is 1.52. What is the focal length of thelens?

Solution

By our convention, the radius of curvature of the first convex surface is +25.0 cm.The inside of the second convex surface is concave to the incident light and itsradius of curvature is therefore −25.0 cm. The focal length of the lens, found by thelensmaker’s formula, equation 31.61, is

1f = (n − 1) 1

R1− 1

R2

= (1.52 − 1) 125 cm − 1

−25.0 cm = 0.0416 cm−1

f = 24.0 cm


Example 31.15

Finding the focal length of a diverging lens. A double concave lens has radii ofcurvature of 25.0 cm. The index of refraction is 1.52. What is the focal length of thelens?


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Solution

By our convention, the radius of curvature of the first concave surface is −25.0 cm.The inside of the second concave surface is convex to the incident ray, and hence,the radius of curvature is +25.0 cm. The focal length of the diverging lens, foundfrom equation 31.61, is

1f = (n − 1) 1

R1− 1

R2

= (1.52 − 1) 1−25.0 − 1

+25.0 = −0.0416 cm−1

f = −24.0 cm

Note that converging lenses have positive focal lengths while diverging lenses havenegative focal lengths.


31.14 Optical InstrumentsMost optical devices consist of an arrangement or combination of lenses. The focallengths of the lenses and their relative positions determines the function of theoptical system. A very brief description of these optical instruments is given here asan example of lens applications. Most optical instruments consist of many lenscombinations to eliminate various lens defects. We will look at only combinations ofsingle lenses in order to demonstrate the principle of the device. More detaileddiscussions can be found in books on optics. The first optical instrument consideredis the camera.

The CameraThe basic elements of a camera, figure 31.28(a), are a converging lens, contained ina bellows or tube type mechanism; a roll of photographic film that can be rolledacross a focal plane; an aperture that determines the amount of light that entersthe camera; and a shutter mechanism that determines the time that the lens isopen to light. A schematic diagram of a camera is shown in figure 31.28(b). A raydiagram shows the location of the image. Because the focal length of the lens is aconstant, increasing or decreasing the object distance p changes the image distanceq, as can easily be determined by the lens equation. Therefore, in order to get asharp image of any object, we need to be able to change the distance from the lens tothe film plane. On studio type cameras, this is accomplished by moving the bellowsinward or outward. On small hand cameras, such as a 35 mm camera, there are twoconcentric tubes containing the lenses. Rotating the tubes clockwise or


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counterclockwise causes the lens to move in or out. The total variation of q isrelatively small for a 35 mm camera.

Figure 31.28 A simple camera.

Example 31.16

The change in the image distance for a 35 mm camera. A 35 mm camera has a focallength of 50.0 mm. By how much will the image distance change whenphotographing one object at infinity and another 1.00 m away?

Solution

When the object is at infinity, the image distance q is equal to the focal length f, asseen in section 31.10, case 1. Therefore,

q1 = f = 50.0 mm = 5.00 cm

When p2 = 1.00 m = 100 cm, the image distance q2 found from the lens equation, is

1 = 1 − 1 = 1 − 1 q2 f p2 5.00 cm 100 cm

q2 = 5.26 cm

The total variation in the image distance q becomes

∆q = q2 − q1 = 5.26 cm − 5.00 cm = 0.26 cm


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Thus, the lens does not have to move very far at all to get the two different pictures.One of the reasons for this is, of course, the relatively small focal length of the lens.If the focal length of the camera had been 20.0 cm, then the same calculation asabove would have given a variation of the image distance ∆q of 5.00 cm. Therefore,to obtain a small variation of q a small focal length is used.


Example 31.17

Getting the entire picture into the camera. At what distance from the 35 mm cameraof example 31.16 should a man, 2.00 m tall, stand in order for his entire length to bein the picture?

Solution

For 35 mm film, the maximum height that the image hi can be is 35.0 mm.Therefore, the magnification of the camera is

M = hi = 3.50 cm = 0.0175 ho 200 cm

Since M = q/p, we can solve for the object distance, p, as

p = q M

Now, as seen above, the image distance q is quite close to the focal length f, so wewill assume that they are equal, that is,

q = fHence,

p = q = f = 5.00 cm = 286 cm = 2.86 m M M 0.0175

That is, for a 2.00-m tall man to have his entire body in the picture he must stand2.86 m from the camera. Note that the magnification of a camera is quite small; thisis necessary because we usually want to put a large scene onto a small piece of film.The images must be greatly reduced if they are to fit on the film.



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http://www.farmingdale.edu/faculty/peter-nolan/pdf/univphys/UPEX31-16,17.XLS

http://www.farmingdale.edu/faculty/peter-nolan/pdf/univphys/UPEX31-16,17.XLS

The amount of light impinging on the film depends on the speed of theshutter, that is, 1/50, 1/100, 1/200, 1/500 of a second and so forth, and the size of theopening at the lens. The size of the opening is determined by an iris diaphragm andis calibrated in terms of the f-number. The f-number is defined as

f# = focal length = f (31.62) diameter of aperture d

The standard f-numbers of a camera are: f/2.8, f/4, f/5.6, f/8, f/11, and f/16. Thus, anf-number of f/4 means that the diameter d of the aperture is 1/4 of the value of thefocal length. If a 50.0-mm focal length is being used, then an f-number of f/4 impliesthat the lens opening is

d = f = f = 50.0 mm = 12.5 mm f# 4 4The opening for f/16 is

d = f = f = 50.0 mm = 3.13 mm f# 16 16

Hence, the larger the f-number of a camera, the smaller the lens opening. Thus, apicture taken in bright sunshine would use a larger f-number than one taken insubdued lighting, for the same shutter speed.

The Simple MicroscopeThe simple microscope is the same as the magnifying glass that was treated insection 31.10, case 6. The object is placed within the principal focus of a converginglens and an enlarged, erect, and virtual image is formed.

The Compound MicroscopeThe compound microscope was invented by Galileo in 1610. It consists of twoconverging lens. The first lens is called the objective lens and is of short focal length.The object to be magnified is placed just outside the focal length f1 of the objective,so that a real, enlarged image is formed at q1, figure 31.29(a). The second lens iscalled the eyepiece, and the image of the objective lens becomes the object for theeyepiece lens. The eyepiece is situated such that the object p2 always lies just withinthe principal focus of the eyepiece f2. Thus the eyepiece acts as a simple magnifyingglass, giving an enlarged virtual image at q2. The lens equation is used to find thelocation of the final image. Thus, q1 is found from

1 = 1 − 1 q1 f1 p1

Once q1 is determined, the object distance p2 is found from


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Figure 31.29 The compound microscope.

p2 = d − q1

The final image is found from 1 = 1 − 1

q2 f2 p2

The magnification of the entire system is

M = M1M2

whereM1 = − q 1 and M2 = − q2

p1 p2

A typical student microscope is shown in figure 31.29(b).

Example 31.18

The microscope. A bug 2.00 mm in diameter is placed 1.50 cm in front of a 1.00-cmfocal length objective lens of a compound microscope. The eyepiece lens has a focallength of 23.0 cm and the lenses are separated by a distance of 25.0 cm. Find thesize of the resulting image.


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Solution

We treat the microscope as a system of two lenses in combination. The image of thefirst lens is the object for the second lens. The image of the first lens, found from thelens equation, equation 31.25, is

1 = 1 − 1 q1 f1 p1

= 1 − 1 1.00 cm 1.50 cm

q1 = 3.00 cm This image serves as the object of the second lens. The object distance for the secondlens, found from equation 31.27, is

p2 = d − q1 = 25.0 cm − 3.00 cm

= 22.0 cm

The image of the second lens, found from the lens equation, equation 31.26, is

1 = 1 − 1 q2 f2 p2

= 1 − 1 23.0 cm 22.0 cm

q2 = −506 cm

The magnification of the first lens is found as

M1 = − q1 = −3.00 cm = −2.00 p1 1.50 cm

whereas the magnification of the second lens is found as

= +23.0M2 = −q2p2

= − −506 cm22.0 cm

The total magnification of the system, found from equation 31.33, is

M = M1M2 = (2.00)(23.0) = 46.0

The size of the final image, found from equation 31.34, is

hi2 = Mho1 = (46.0)(2.00 mm)= 92.0 mm = 9.20 cm


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Notice that because M1 is negative, the first image is real, whereas M2 is positive,indicating that the second image is virtual. This is as expected, since the object p2

lies within the principal focus f2 of the second lens. The final result of the opticalsystem is to take the initial object, 2.00 mm in size, and make it observable as animage 92.0 mm in size.


The Astronomical TelescopeThe astronomical telescope is another example of a combination of two converginglenses. However, this time the objective is a large converging lens of a long focallength. The astronomical telescope is shown in figure 31.30(a). Because the object isvery far away, the image of the objective is, for all intents and purposes, located atthe principal focus of the objective lens fo. (Recall that when p = ∞, q = f.) It is shownslightly away from there in figure 31.30 so that the ray diagram is easier to see. Theeyepiece is a converging lens of small focal length, fe, and it is placed such that theimage of the first lens falls just within the principal focus of the second lens, therebyforming an enlarged, virtual image. That is, the eyepiece acts as a simplemagnifying glass. Since the first image falls approximately at the focal distance ofthe first lens, and the second lens is placed so that the image of the first lens fallsjust within the focal length of the second lens, the distance separating the lenses isapproximately the sum of the two focal lengths. Hence, the overall length of thetelescope is

L = fo + fe (31.63)

The final image of the astronomical telescope is, of course, inverted but this isnot serious for analyzing the heavenly bodies. However, the astronomical telescopeis not very good for viewing distant objects on the surface of the earth, because theyare all inverted. (The astronomical telescope can be converted to a terrestrialtelescope by placing a third converging lens between the objective and the eyepiece.)The magnification of a telescope is usually expressed in terms of the angularmagnification. The angular magnification MA of the lens is defined as the ratio ofthe angle subtended at the eye by the object when the lens is used, to the anglesubtended by the unaided eye. We state without proof that the total angularmagnification of an astronomical telescope when the object is at infinity is given by

MA = focal length of objective = fo (31.64) focal length of eyepiece fe


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(a)

(b)Figure 31.30 The astronomical telescope.

Example 31.19

An astronomical telescope. Find the angular magnification of an astronomicaltelescope whose objective has a focal length of 100 cm, and its eyepiece has a focallength of 5.00 cm.

Solution

The magnification, found from equation 31.48 is,


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M = fo = 100 cm = 20.0 fe 5.00 cm


The Galilean TelescopeThe Galilean telescope consists of a converging lens of large focal length for theobjective and a diverging lens of short focal length for the eyepiece. The advantageof the Galilean telescope is that it gives an erect image. A ray diagram for aGalilean telescope is shown in figure 31.31. Only two of the standard rays are

Figure 31.31 Ray diagram for a Galilean telescope.

shown to simplify the diagram. The image of the first lens is located at q1. But notethat the second lens is placed before the location of the first image. This is anexample of what is called a virtual object, and the object distance p must be writtenas a negative quantity in the lens equation. The effect of the diverging lens for theeyepiece is to diverge the rays away from the position where they should focus bythe objective lens. For example, when ray (2) hits the diverging lens it is parallel tothe principal axis. However, a ray parallel to the principal axis is diverged awayfrom the principal axis as though it came from the principal focus of the diverginglens. This ray is now shown as diverging away from the principal axis. In thisparticular example, the second lens is located at the principal focus of the first lens,hence ray (1) passes through the geometrical center of lens two and is not deviated.We can see from the figure that these two rays diverge on the right-hand side of thesecond lens and therefore can never form a real image there. However if these raysare produced backward they intersect on the left-hand side of the lens to form anenlarged, erect, virtual image. Thus the Galilean telescope, also called a terrestrialtelescope, produces enlarged images that are right side up.


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Example 31.20

A Galilean telescope. A Galilean telescope is made of a + 20.0-cm objective lens anda −4.50-cm eyepiece lens. The two lenses are separated by a distance d of 15.0 cm.For an object situated at 1000 cm, find (a) the image distance of the first lens,(b) the object distance for the second lens, (c) the image distance for the second lens.

Solution

a. The image distance of the first lens, found from the lens equation, equation 31.25,is

1 = 1 − 1 = 1 − 1 q1 f1 p1 20.0 cm 1000 cm

q1 = 20.4 cm

b. The object distance for the second lens, found from equation 31.27, is

p2 = d − q1 = 15.0 cm −20.4 cm = −5.4 cm

Note that this is a virtual object since p2 is negative.

c. The image distance for the second lens, found from equation 31.26, is

1 = 1 − 1 = 1 − 1 q2 f2 p2 (−4.50 cm) (−5.4 cm)

q2 = −27 cm

Because q2 is negative the final image is virtual.


Have you ever wondered…?An Essay on the Application of Physics

Nature’s Camera -- the Human Eye

Have you ever wondered how the human eye works? It is one of the greatestmarvels of nature, but one that is usually taken for granted. The human eye is verymuch like a camera. It contains a lens, an iris diaphragm to let the light in, and itsfilm is the retina at the back of the eye, figure 1. The lens focuses incident light onto


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the retina so a sharp image is observed. The iris diaphragm opens wide in subduedlighting and closes down in very bright light.

In the eye, the distance from the lens to the retina q, does not change as in acamera, instead the shape of the lens is changed by the muscles and ligaments of

Figure 1 The human eye.

the eye. Hence, the eye changes its focal length in order to bring objects at differentobject distances to a focus at the same image distance q, the distance from the lens tothe retina. This changing of the focal length of the eye is called accommodation. Aschematic of the normal eye is shown in figure 2(a). The greatest amount of bending

Figure 2 Defects of the human eye.

of light in the eye occurs at the cornea, because it is here where the greatestdifference in the index of refraction occurs. The index of refraction of air is of courseequal to one, whereas the index of refraction of the cornea is about 1.35. The indexof refraction of the lens is about 1.44; that of the aqueous and vitreous humor isabout 1.34.


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Some eyes do not have the ability to change the shape of the lens asnecessary and this is referred to as a defect in vision. One such defect is calledhyperopia, or farsightedness, and is shown in figure 2(b). A farsighted personcan see distant objects clearly, but objects a short distance away are blurred. The lensis not capable of bending the incoming light of near objects to a focus on the retina.Instead the light is focused behind the retina, leaving blurred vision. The defect iseasily remedied by placing a converging lens in front of the eye, as shown in figure2(c). The converging lens brings these incoming rays to a focus at the retina.Farsightedness is a defect that occurs to most people as they age. The lens hardenswith age and the muscles weaken and are no longer capable of shaping the lens asneeded.

Another defect is myopia, or nearsightedness, as illustrated in figure 2(d).A nearsighted person can see nearby objects clearly, but distant objects are blurred.The distant objects come to a focus in front of the retina leaving a blurred image onthe retina. This defect can be eliminated by placing a diverging lens in front of theeye. The diverging lens causes the rays to converge more slowly and hence theimage is formed farther back at the retina, as shown in figure 2(e).

The minimum distance from the eye at which an object can be seen distinctlyis called the near point of the eye. For the average person, the near point is about25.0 cm. That is, an object at a distance p of 25.0 cm is the minimum distance thatan object can be placed to give a distinct image on the retina. If the distance isdecreased below this value, the image will be blurred for most people. Remember,this is only an average value, as we get older, the minimum distance of distinctvision increases. For example, the near point can be as close as 6.00 cm for a childand can increase to over a 100 cm for an elderly person.

Example 31H.1

Range of focal length for the eye. The distance from the cornea to the retina in thehuman eye is about 20.0 mm. What should the focal length of the human eye be inorder to see clearly an object (a) at infinity and (b) at the near point of most distinctvision, 25 cm?

Solution

a. For the object at infinity, the image is located at the principal focus. Hence

f = q = 20.0 mm = 2.00 cm

is the focal length of the eye when it views an object at infinity.

b. For the object at p = 25.0 cm, the focal length would have to be


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1 = 1 + 1 = 1 + 1 = 0.540 cm−1

f p q 25.0 cm 2.00 cmf = 1.85 cm

Thus, the eye only has to change its focal length by (2.00 cm − 1.85 cm = 0.15 cm) 15mm to see its entire range of vision.

Another common defect of the eye is called astigmatism. Astigmatism occursin some people, because their eye is not completely spherical. That is, the radius ofcurvature of the eye in the vertical direction is not the same as the radius ofcurvature in the horizontal direction. Hence, vertical rays do not converge to thesame position as horizontal rays. This defect is usually corrected with a lens ofcylindrical curvature.

Although the magnification of a lens was defined in equation 31.22 as M =hi/ho, when viewing an object with the eye it is sometimes more desirable to talkabout the angular magnification of the lens. Figure 3(a) shows an object at threedistances from the unaided eye. The size of the object, as determined by the unaided

Figure 3 Angular magnification and the simple microscope.

eye, depends on the angle subtended by the object at the eye. Thus, when the objectis at position 1, it subtends an angle θ1; when it is at position 2, it subtends theangle θ2; and when it is at position 3, it subtends the angle θ3. As seen from thediagram, θ1 < θ2 < θ3. Thus, the closer the object is to the eye, the greater the angle


31-56

subtended and the easier it is for the eye to see the object. However, the objectcannot be moved closer to the eye than the near point, because that is the closestpoint that the eye can still see the object distinctly. In figure 3(b), the object isplaced at the object distance p1 = 25.0 cm in front of a normal eye. The height of theobject ho subtends an angle α at the lens of the eye. The image of the object isfocused on the retina of the eye. This is the largest image that the unaided eye canmake of the object. If the object is placed within the focal point of a converging lens,the object can be brought even closer to the eye and hence will subtend a largerangle β, figure 3(c), than with the unaided eye. Therefore, the object is placed withinthe focal point, at the position p2, such that the image distance q2 is equal to thenear point of the eye, the closest position of distinct vision. We assume that thevalue of the near point q2 = 25.0 cm. The image is virtual, erect, and magnified asseen earlier. The angular magnification MA of the lens is defined as the ratio ofthe angle subtended at the eye by the object when the lens is used, to the anglesubtended by the unaided eye. That is,

MA = β (31H.1) α

From figure 3(c), we see thattan β = ho p2

whereas from figure 3(b), we see that

tan α = ho = ho p1 25.0 cmLet us take the ratio

tan β = ho/p2 tan α ho/25.0 cm

= 25.0 cm p2

Because the angles considered are small, we can use the small-angle approximation,setting the tangent of an angle equal to the angle itself. Hence,

β = 25.0 cm α p2

But this ratio is the definition of the angular magnification, equation 31H.1, hence

MA = 25.0 cm (31H.2) p2

The value of p2, found from the lens equation, is


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1 = 1 − 1 p2 f q2

1 = 1 − 1 = 25.0 cm + f (31H.3) p2 f −25.0 cm (25.0 cm)f

And the object distance for the image to be at the near point is

p2 = (25.0 cm)f (31H.4) 25.0 cm + f

Substituting equation 31H.4 into equation 31H.2, gives

MA = β = 25.0 cm = 25.0 cm α p2 (25.0 cm)f /(25.0 cm + f)

MA = 25.0 cm + 1 (31H.5) f

Equation 31H.5 gives the angular magnification of a converging lens when the imageis located at the near point of the eye, which is assumed to be 25.0 cm for the normaleye.

Example 31H.2

The angular magnification of a lens when the image is at the near point of the eye. Aconverging lens of 10.0-cm focal length is used as a magnifying glass. If the image isto be at the near point of the eye, find the angular magnification of the lens.

Solution

The angular magnification, found from equation 31H.5, is

MA = 25.0 cm + 1 f

= 25.0 cm + 1 10.0 cm

= 3.5

The eye is more relaxed when viewing an object at infinity than at the nearpoint. It is, therefore, sometimes more desirable to view an image at infinity than atthe near point. For the image to be at infinity, the object must be placed at theprincipal focus. The lens equation, equation 31H.3, shows that if p is at infinity,


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then the object distance p2 is equal to the focal length f. Hence substituting f for p2

in equation 31H.2, givesMA = 25.0 cm (31H.6)

f

Equation 31H.6 gives the angular magnification of a simple microscope when theimage is set for viewing at infinity.

Example 31H.3

The angular magnification of a lens when the image is at infinity. A magnifyingglass of 10.0-cm focal length is used to view an object, such that the image is atinfinity. Find the angular magnification of the lens.

Solution

The angular magnification, found from equation 31H.6, is

MA = 25.0 cm f

= 25.0 cm = 2.50 10.0 cm

Notice that when the image is viewed at the near point the angularmagnification is 3.5, whereas when viewed at infinity, it is 2.5. If viewing is for along time, it might be desirable to view the image at infinity because it is easier onthe eyes. If the magnification is the most important consideration, then the imageshould be viewed at the near point of the eye.

The Language of Physics

RefractionThe bending of light as it travels from one medium into another. It occurs because ofthe difference in the speed of light in the different mediums. Whenever a ray of lightgoes from a rarer medium to a denser medium the refracted ray is always benttoward the normal. Whenever a ray of light goes from a denser medium to a rarermedium, the refracted ray is bent away from the normal (p. ).


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Law of refractionThe ratio of the sine of the angle of incidence to the sine of the angle of refraction isa constant. The constant is called the relative index of refraction and it is equal tothe ratio of the speed of light in the first medium to the speed of light in the secondmedium. Because of the changing speed of light, the wavelength of light changes asthe light passes into the second medium (p. ).

The critical angle of incidenceThe angle of incidence that causes the refracted ray to bend through 900. When theincident angle exceeds the critical angle no refraction occurs. In that case, it iscalled total internal reflection because all the light that strikes the interface isreflected (p. ).

PrismA triangular piece of transparent material whose angle exceeds the critical angle. Aray of light falling on one of the smaller sides of the prism enters the prism and istotally reflected from the longer side of the prism. Prisms are also used foranalyzing the dispersion of white light into its component colors (p. ).

Fiber opticsA flexible glass rod of high refractive index. Light entering the glass undergoes totalinternal reflection from the walls of the glass fiber and the light travels down thelength of the fiber with little or no absorption of light (p. ).

DispersionThe separation of white light into its component colors. It occurs because the indexof refraction of a medium varies slightly with the wavelength of light (p. ).

LensA piece of transparent material, such as glass or plastic, that causes light passingthrough it to either converge or diverge depending on the shape of the material (p. ).

Lensmaker’s formulaAn equation that relates the focal length, index of refraction of the medium and thelens, and the radii of curvature of the lens (p. ).

Thin lensA lens whose thickness is negligible compared to the distance to the principal focusand to any object or image distance (p. ).

Converging lensA lens that causes light, parallel to its principal axis, to converge to the principalaxis. Converging lenses have positive focal lengths (p. ).


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Diverging lensA lens that causes light, parallel to its principal axis to diverge away from theprincipal axis. Diverging lenses have negative focal lengths (p. ).

The lens equationAn equation that relates the image distance, the object distance, and the focallength of a lens. It has the same form as the mirror equation (p. ).

Dioptric power of a lensThe reciprocal of the focal length of a lens. The focal length must be expressed inmeters. For a combination of any number of lenses in contact, the power of thecombination is equal to the sum of the powers of each individual lens (p. ).

AccommodationThe changing of the focal length of the eye in order to focus an image on the retinaof the eye (p. ).

Hyperopia, or farsightednessA defect of the eye that causes objects far away to be seen clearly while close objectsare blurred. The condition is remedied by placing a converging lens in front of theeye (p. ).

Myopia, or nearsightednessA defect of the eye that causes objects close to the eye to be seen clearly, whileobjects far away are blurred. The condition is remedied by placing a diverging lensin front of the eye (p. ).

Near point of the eyeThe minimum distance from the eye at which an object can be seen distinctly. Forthe average person the near point is about 25 cm (p. ).

Angular magnificationThe ratio of the angle subtended at the eye by an object, when a lens is used, to theangle subtended by the unaided eye (p. ).

Summary of Important Equations

The law of refraction sin i = v1 = constant = n21 (31.3) sin r v2

n1 sin i = n2 sin r (31.8)The index of refraction n = c (31.4)

v


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Speed of a wave in terms of wavelength and frequency v = λν (31.9)

Index of refraction in terms of wavelengths in two mediums n21 = λ1 (31.10) λ2

Apparent depth for small angles q = n2 p (31.14) n1

Critical angle sin ic = n2 (31.17) n1

Lens equation 1 = 1 + 1 (31.21) f p qMagnification M = hi = − q (31.23) ho p Height of image hi = Mho (31.24)

Combination of lensesFinal image 1 = 1 − 1 (31.26) q2 f2 p2 where p2 = d − q1 (31.27)and 1 = 1 − 1 (31.25) q1 f1 p1

Magnification M = M1M2 (31.33)

Thin lenses in contact 1 = 1 + 1 + 1 + 1 + … (31.43) fc f1 f2 f3 f4

Power of a lens P = 1 (31.44) f (in meters)

Power of combinations of lenses Pc = P1 + P2 + P3+ P4 + (31.45)

Lensmaker’s formula, lens in any medium (31.60)1f = n2 − n1

n11

R1− 1

R2

Lensmaker’s formula, lens in air (31.61)1f = (n − 1) 1

R1− 1

R2

f-number of a lens f# = f (31.62) d

Angular magnification, astronomical telescope MA = fo (31.64) fe

Angular magnification, viewing at near point MA = 25.0 cm + 1 (31H.5) fAngular magnification, viewing at infinity MA = 25.0 cm (31H.6)

f


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Questions for Chapter 31

*1. Why does a diamond sparkle?2. When the angle of incidence is equal to zero, the angle of refraction is also

zero. Does the wavelength of the light change when going from one medium toanother under these circumstances?

3. If you are at the bottom of a pool of water and you look upward into the airat an angle greater than 500 can you see anything in the air?

4. Is it possible to have a ray of light refracted into a medium such that thenew wavelength decreases to the point where it is no longer in the visible spectrum?What would you see? Does the eye interpret a wavelength or a frequency?

5. What does a wide-angle lens and a telephoto lens do when each is attachedto a camera?

6. Describe the optical system used for (a) a slide projector, (b) a moviecamera, and (c) an overhead projector.

7. A swimmer forgets to take off her glasses as she enters the pool. When sheis under water, wearing the glasses, does the water have any effect on what she cansee with the glasses?

8. When you see yourself in a mirror, your right and left hand areinterchanged. Yet, if you see a picture of yourself made with a camera, containinglenses, your right and left are not changed. Why?

*9. What are bifocals and why are they used?*10. What is a mirage and how is it explained by the index of refraction?*11. How is a rainbow formed?12. How is a convex lens used, with the help of the sun, to start a fire?

Problems for Chapter 31

31.2 The Law of Refraction1. A ray of light impinges on a piece of glass (ng = 1.52) at an angle of

incidence of 50.00. Find the angle of refraction.2. A ray of light passes from water to glass at an angle of incidence of 50.00.

Find the angle of refraction.3. A ray of light is refracted by an angle of 34.50 as it enters water from glass.

Find the angle of incidence.4. A ray of light in air makes an angle of incidence of 35.00 as it enters an

unknown liquid. The refracted ray in the fluid is measured to be 22.50. Find theindex of refraction of the unknown liquid. What substance might it be?


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Diagram for problem 4.

5. Determine the speed of light in (a) water, (b) glycerine, and (c) diamond.6. If a ray of light of 480.0-nm wavelength in air enters into water, what is

the wavelength of the light in the water? Is this light visible?7. A ray of light of 580.0-nm wavelength in water, enters into the air. What is

the wavelength of the light in the air?8. A ray of light of 700.0-nm wavelength in water enters into the air. What is

the wavelength of the light in the air? Will the light be seen in the air?9. A ray of light of 590.0-nm wavelength in air impinges on a piece of flint

glass at an angle of 30.00 with the vertical. The index of refraction of this flint glassis 1.57. Find (a) the angle of refraction, (b) the speed of light in the glass, and (c) thewavelength of light in the glass.

31.3 Apparent Depth of an Object Immersed in Water10. A rock sits at the bottom of a 3.50-m-deep pool. What is its apparent

depth?11. A fish appears to be at a depth of 120 cm in water. What is its actual

depth?


31.4 Refraction through Parallel Faces12. A ray of light in air makes an angle of incidence of 30.00 with a sheet of

glass 0.500 cm thick. Find the distance d that the final ray is displaced from itsoriginal direction.


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31.5 Total Internal Reflection13. What is the critical angle of refraction for a light ray going from water to

glass?14. Find the critical angle of refraction for a ray of light passing from

glycerine (n = 1.47) into air.15. The critical angle of a ray of light is measured as 41.80 as it goes from an

unknown liquid into air. Find the index of refraction of the liquid.

31.7 Thin Lenses16. A double convex glass lens of index of refraction 1.52, has radii of

curvature R1 = +50.0 cm and R2 = −25.0 cm. Find its focal length.17. A glass lens of index of refraction 1.52, has radii of curvature of R1 =

+40.0 cm and R2 = −15.0 cm. What is its focal length and is it a converging or adiverging lens?

18. A glass lens of index of refraction 1.52, has radii of curvature of R1 = −40.0cm and R2 = +15.0 cm. What is its focal length and is it a converging or a diverginglens?

19. A glass lens of index of refraction 1.52, has radii of curvature of R1 = −40.0cm and R2 = −15.0 cm. What is its focal length and is it a converging or a diverginglens?

20. A lens is made of transparent material, with radii of curvature R1 = 15.0cm and R2 = 90.0 cm. If the focal length of the lens is 5.00 cm, find the index ofrefraction of the material.

31.9 The Lens Equation, and Section 31.10 Some Special Cases for theConvex Lens

*21. An object 3.00 cm high is placed 20.0 cm in front of a converging lens of15.0 cm focal length. Draw a ray diagram. Find (a) the image distance, (b) the


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magnification, and (c) the final size of the image. (d) Is the image real or virtual?(e) Is the image erect or inverted?

*22. An object 3.00 cm high is placed 20.0 cm in front of a diverging lens of−15.0 cm focal length. Draw a ray diagram. Find (a) the image distance q, (b) themagnification M, and (c) the height of the image hi. (d) Is the image real or virtual?(e) Is the image erect or inverted?

23. Where should an object be placed in front of a 20.0-cm lens in order forthe image to be the same size as the object?

24. An object 7.00 cm high is placed 5.00 cm in front of a convex lens of 10.0cm focal length. (a) Draw a ray diagram. Find (b) the image distance, (c) themagnification, and (d) the height of the image.

25. How far in front of a 20.0-cm converging lens should an object be placed inorder to produce an image 25.0 cm from the lens on the same side of the lens as theobject?

26. An object is placed 15.0 cm in front of a diverging lens of 5.00 cm focallength. Where is the image located and what is its magnification?

27. Where should an object be placed in front of a diverging lens of 10.0 cmfocal length in order to give an image a magnification of 0.500?

28. An object 5.00 cm high is placed 20.0 cm in front of a converging lens. Theimage is measured to be 7.00 cm high. Where is the image located?

29. An object 5.00 cm high is placed 30.0 cm in front of a converging lens of7.50 cm focal length. What is the size of the image?

31.11 Combination of Lenses*30. An object 5.00 cm high is placed 20.0 cm in front of a converging lens of

10.0 cm focal length. A second converging lens of 20.0 cm focal length is placed 30.0cm behind the first lens. Find (a) the image distance for the first lens, (b) the objectdistance for the second lens, (c) the image distance for the second lens, and (d) thetotal magnification of the combination. Draw a ray diagram.

*31. An object 5.00 cm high is placed 10.0 cm in front of a 20.0-cm converginglens. A second converging lens, also of 20.0 cm focal length, is placed 20.0 cm behindthe first lens. Find (a) the location of the final image and (b) its size. Draw a raydiagram.

*32. A converging lens of +20.0 cm is separated by a distance of 20.0 cm froma diverging lens of −5.00 cm. An object is located 30.0 cm in front of the first lens.Find (a) the image distance of the first lens, (b) the object distance for the secondlens, (c) the image distance for the second lens, and (d) the magnification of thesystem.

31.12 Thin Lenses in Contact33. A 20.0-cm convex lens is placed in contact with a diverging lens of

unknown focal length. The lens combination has a focal length of 30.0 cm. Find thefocal length of the diverging lens.


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34. What is the power of (a) a 50.0-cm converging lens, (b) a 20.0-cmdiverging lens, (c) the converging and diverging lenses in contact, and (d) the focallength of the combination when they are in contact?

35. Ten identical converging thin lenses, each of focal length 6.50 cm, are incontact. Find the focal length of the composite lens.

Additional Problems36. A ray of light of 590.0-nm wavelength in glycerine impinges on a piece of

flint glass at an angle of 30.00 with the vertical. The index of refraction of glycerineis 1.47 and for the flint glass used it is 1.70. Find (a) the angle of refraction, (b) thespeed of light in the glycerine, (c) the speed of light in the glass, and (d) thewavelength of light in the glass.

37. Light travels in medium 1 at 1.50 × 108 m/s. The light is incident at anangle of incidence of 40.00 when entering medium 2, where the angle of refraction is300. Find the speed of light in medium 2.

38. A ray of light of 590.0-nm wavelength makes an angle of incidence of40.00 on to the cornea of the eye. The index of refraction of the cornea is 1.35. Find(a) the angle of refraction, (b) the speed of light in the cornea, and (c) thewavelength of light in the cornea.

39. Light travels from glass of index of refraction 1.50 into water of index1.33. If there exists a refracted ray, find the angle of refraction when (a) the angle ofincidence is 55.00 and (b) the angle of incidence is 70.00.

*40. If an object is placed a distance d1 in front of the principal focus of a lensand the image is located a distance d2 beyond the other principal focus, show thatthe focal length of the lens is given by

f = d1d2

41. An object is mounted 60.0 cm in front of a screen. At what two positionswill a 12.0 cm focal length lens yield a distinct image on the screen?

42. An optical system creates a virtual object (an object behind the lens) for adiverging lens of focal length f = −8.00 cm. Find the position of the image anddescribe it for the following two object distances: (a) p = −4.00 cm and (b) p = −10.0cm.


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43. A farsighted person has a near point of 60.0 cm. What power lens shouldbe used for eyeglasses such that the person can read this book at a distance of 25.0cm?

44. A nearsighted person has a far point of 15 cm. (The far point of the eye isthe farthest distance that an object can be seen clearly.) What power lens should beused to allow this person to view far distant objects?

45. What is the smallest wavelength in air of visible light that can still beseen on refraction into water?

46. A glass lens, ng = 1.70, has a focal length of 25.0 cm in air. What is itsfocal length when it is submerged in water?

Interactive Tutorials47. The law of refraction. A ray of light of wavelength λ1 = 500.0 nm in

medium 1 (index of refraction of medium one is n1 = 1.00) impinges on a secondmedium of index of refraction n2 = 1.52, at an angle of incidence i = 35.00. Find(a) the angle of refraction r, (b) the speed of light v2 in the second medium, and(c) the wavelength λ2 in the second medium.

48. The critical angle of incidence. A ray of light in the denser medium (indexof refraction n1 = 1.50) impinges on a less dense medium of index of refraction n2 =1.00. Find the critical angle of incidence ic for the ray of light going from the moredense medium to the less dense medium.

49. Lensmaker’s formula. An optical lens has an index of refraction n = 1.52and radii of curvature of R1 = +35.0 cm and R2 = −15.0 cm. Find the focal length ofthis lens.

50. Thin lens. An object of height ho = 8.50 cm is placed at the object distancep = 35.0 cm of a thin lens of focal length f = 15.0 cm. Find (a) the image distance q,(b) the magnification M, and (c) the height hi of the resulting image.

51. Thin lens. An object of height ho = 0.500 cm is placed at the objectdistance p = 2.00 cm of a thin lens of focal length f = 0.700 cm. Plot (a) the imagedistance q and (b) the height hi of the resulting image as the object is moved towardthe lens.

52. A combination of lenses. An object of height ho1 = 5.00 cm is placed at theobject distance p1 = 15.0 cm of a thin lens of focal length f1 = 10.0 cm. A second lensof focal length f2 = 10.0 cm is placed d = 48.0 cm behind the first lens. Find (a) theimage distance q1 of the first lens, (b) the magnification M1 of the first lens, (c) theobject distance p2 of the second lens, (d) the image distance q2 of the second lens,(e) the magnification M2 of the second lens, (f) the magnification M of the system,and (g) the height of the final image hi2.

To go to these Interactive Tutorials click on this sentence.

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Chapter 31 The Law of Refraction