Optics Review #1

LCHSDr.E

When a light wave enters a new medium and is refracted, there must be a change in the light wave’s

(A) color (B) frequency (C) period (D) speed

In a vacuum, all electromagnetic waves have the same

(A) speed (B) phase (C) frequency(D) wavelength

The diagram represents a light ray reflecting from a plane mirror. The angle

of reflection for the light ray is

(A) 25° (B) 35° (C) 50° (D) 65°

In which way does blue light change as it travels from diamond into glass?

(A) Its frequency decreases. (B) Its frequency increases. (C) Its speed decreases. (D) Its speed increases.

Which phenomenon provides evidence that light has a wave nature?

(A) emission of light from an energy- level transition in a hydrogen atom(B) diffraction of light passing through a narrow opening(C) absorption of light by a black sheet of paper (D) reflection of light from a mirror

Which form(s) of energy can be transmitted through a vacuum?

(A) light, only (B) sound, only (C) both light and sound (D) neither light nor sound

The diagram represents a light ray striking the boundary between air and glass. What would be the angle between this light ray and its reflected ray?(A) 30° (B) 60° (C) 120° (D) 150°

Which quantity is equivalent to the product of the absolute index of refraction of water and the speed of light in water?

(A) wavelength of light in a vacuum (B) frequency of light in water (C) sine of the angle of incidence (D) speed of light in a vacuum

What happens to the frequency and the speed of an electromagnetic wave as it passes from air into glass?

(A) The frequency decreases and the speed increases.(B) The frequency increases and the speed decreases. (C) The frequency remains the same and the speed increases.(D) The frequency remains the same and the speed decreases.

Which ray diagram best represents the phenomenon of refraction?

Mirror

The +/- Sign ConventionsThe sign conventions for the given quantities in the mirror equation and magnification equations are as follows:f is + if the mirror is a concave mirrorf is - if the mirror is a convex mirrordi is + if the image is a real image and located on the object's side of the mirror.di is - if the image is a virtual image and located behind the mirror.hi is + if the image is an upright image (and therefore, also virtual)hi is - if the image an inverted image (and therefore, also real)

A 4.0-cm tall light bulb is placed a distance of 8.3 cm from a concave mirror having a focal length of 15.2 cm. Determine the image distance and the image size.1/f = 1/do + 1/di1/(15.2 cm) = 1/(8.3 cm) + 1/di

0.0658 cm-1 = 0.120 cm-1 + 1/di

-0.0547 cm-1 = 1/di

18.3 = = di

hi/ho = - di/do

hi /(4.0 cm) = - (-18.2 cm)/(8.3 cm)hi = - (4.0 cm) • (-18.2 cm)/(8.3 cm)hi = 8.8 cm

Determine the image distance and image height for a 5.00-cm tall object placed 45.0 cm from a concave mirror having a focal length of 15.0 cm. di = 22.5 cm and hi = -2.5 cmUse 1/f = 1/do + 1/di where f = 15 cm and do = 45 cmThen use hi / ho = - di / do where ho = 5 cm, do = 45 cm, and di = 22.5 cm

Determine the image distance and image height for a 5.00-cm tall object placed 30.0 cm from a concave mirror having a focal length of 15.0 cm. di = 30.0 cmand hi = -5.0 cmUse 1 / f = 1 / do + 1 / di where f =15 cm and do = 45 cmThen use hi / ho = - di / do where ho = 5 cm, do = 45 cm, and di = 30.0 cm

Determine the image distance and image height for a 5.00-cm tall object placed 20.0 cm from a concave mirror having a focal length of 15.0 cm. di = 60.0 cm and hi = -15.0 cmUse 1 / f = 1 / do + 1/di where f =15 cm and do=45 cmThen use hi / ho = -di / do where ho = 5 cm, do = 45 cm, and di = 60.0 cm

Determine the image distance and image height for a 5.00-cm tall object placed 10.0 cm from a concave mirror having a focal length of 15.0 cm. di = -30.0 cm and hi = +15.0 cmUse 1 / f = 1 / do + 1 / di where f = 15 cm and do = 10.0 cmThen use hi / ho = -di / do where ho = 5 cm, do = 45 cm, and di = -30.0 cm

A magnified, inverted image is located a distance of 32.0 cm from a concave mirror with a focal length of 12.0 cm. Determine the object distance and tell whether the image is real or virtual. di = 19.2 cm and RealUse the equation 1 / f = 1 / do + 1 / diwhere f = 12 cm and do = 32 cmAll inverted images are real images.

An inverted image is magnified by 2 when the object is placed 22 cm in front of a concave mirror. Determine the image distance and the focal length of the mirror.

di = 44 cmand f = 14.7 cm andRealUse the equation 1 / f = 1 / do + 1 / diwhere do = 22 cm and M = 2If M=2 and the image is inverted, then the di must be +44 cm. (It isn't -44 cm since the negative sign would only correspond to an upright and virtual image.)Solve for f one you find di =+44 cmYou know that the image is real if it is described as being upright.

Snell’s Law

Determine the angle of refraction

Answer: 53.9 degreesMeasure the angle of incidence - the angle between the normal and incident ray. It is approximately 45 degrees.Given:ni = 1.52 nr = 1.33theta i = 45 degreesFind: theta rSubstitute into Snell's law equation and perform the necessary algebraic operations to solve:1.52 • sine(45 degrees) = 1.33 • sine (theta r) 1.075 = 1.33* sine (theta r)0.8081 = sine (theta r)53.9 degrees = theta r

Determine the angle of refractionAnswer: 28.4 degreesMeasure the angle of incidence - the angle between the normal and incident ray. It is approximately 60 degrees.Given:ni =1.33 nr = 2.42theta i = 60 degreesFind: theta rSubstitute into Snell's law equation and perform the necessary algebraic operations to solve:1.33 • sine(60 degrees) = 2.42 • sine (theta r) 1.152 = 2.42 • sine (theta r)0.4760 = sine (theta r)28.4 degrees = theta r

Snell’s LawFirst, draw normal and measure the angle of incidence at first boundary; it is approximately 30 degrees. Then, use the given n values and Snell's Law to

calculate the theta r values at each boundary. The angle of refraction at one boundary becomes the angle of incidence at the next boundary; e.g., the theta r at the air-flint glass boundary is the theta i at the flint glass-water boundary. Here are the calculated theta r values:

air - flint glass: 18 degrees flint glass - water: 22 degreeswater - diamond: 12 degrees

diamond - zirconium: 13 degreescubic zirconium - air: 30 degrees

Snell’s LawFirst, draw normal and measure the angle of incidence at first boundary; it is approximately 30 degrees. Then, use the given n values and Snell's Law to

calculate the theta r values at each boundary. The angle of refraction at one boundary becomes the angle of incidence at the next boundary; e.g., the theta r at the air-flint glass boundary is the theta i at the flint glass-water boundary. Here are the calculated theta r values:

air - flint glass: 18 degrees flint glass - water: 22 degreeswater - diamond: 12 degrees

diamond - zirconium: 13 degreescubic zirconium - air: 30 degrees

Critical Angle

Critical Angle = sin-1 (nr/ni)

Calculate the critical angle for the crown glass-air boundary.

The solution to the problem involves the use of the above equation for the critical angle. Critical Angle = sin-1 (nr/ni) = sin-1 (nr/ni)

= sin-1 (1.000/1.52) = 41.1 degrees

Calculate the critical angle for the diamond-air boundary.Critical Angle = sin-1 (1.000/2.42) = 24.4 degrees

Snell’s LawFirst, draw normal and measure the angle of incidence at first boundary; it is approximately 30 degrees. Then, use the given n values and Snell's Law to

calculate the theta r values at each boundary. The angle of refraction at one boundary becomes the angle of incidence at the next boundary; e.g., the theta r at the air-flint glass boundary is the theta i at the flint glass-water boundary. Here are the calculated theta r values:

air - flint glass: 18 degrees flint glass - water: 22 degreeswater - diamond: 12 degrees

diamond - zirconium: 13 degreescubic zirconium - air: 30 degrees

Calculate the critical angle for an ethanol-air boundary. Crit. Angle = sine-1 (ni / nr)Crit. Angle = sine-1 (1.0 / 1.36)Crit. Angle = 47.3 degrees

Calculate the critical angle for an ethanol-air boundary. Crit. Angle = sine-1 (ni / nr)Crit. Angle = sine-1 (1.0 / 1.36)Crit. Angle = 47.3 degrees

Calculate the critical angle for a flint glass-air boundary. Crit. Angle = sine-

1 (ni / nr)Crit. Angle = sine-1 (1.0 / 1.58)Crit. Angle = 39.3 degrees

Calculate the critical angle for a flint glass-air boundary. Crit. Angle = sine-

1 (ni / nr)Crit. Angle = sine-1 (1.0 / 1.58)Crit. Angle = 39.3 degrees

Lenses

33

A 4.00-cm tall light bulb is placed a distance of 45.7 cm from a double convex lens having a focal length of 15.2 cm. Determine the image distance and the image size. 1/f = 1/do + 1/di1/(15.2 cm) = 1/(45.7 cm) + 1/di

0.0658 cm-1 = 0.0219 cm-1 + 1/di

0.0439 cm-1 = 1/di

22.8 cm = di

hi/ho = - di/dohi /(4.00 cm) = - (22.8 cm)/(45.7 cm)hi = - (4.00 cm) • (22.8 cm)/(45.7 cm)= -1.99 cm

A 4.00-cm tall light bulb is placed a distance of 45.7 cm from a double convex lens having a focal length of 15.2 cm. Determine the image distance and the image size. 1/f = 1/do + 1/di1/(15.2 cm) = 1/(45.7 cm) + 1/di

0.0658 cm-1 = 0.0219 cm-1 + 1/di

0.0439 cm-1 = 1/di

22.8 cm = di

hi/ho = - di/dohi /(4.00 cm) = - (22.8 cm)/(45.7 cm)hi = - (4.00 cm) • (22.8 cm)/(45.7 cm)= -1.99 cm

A 4.00-cm tall light bulb is placed a distance of 8.30 cm from a double convex lens having a focal length of 15.2 cm. Determine the image distance and the image size.1/f = 1/do + 1/di1/(15.2 cm) = 1/(8.30 cm) + 1/di

0.0658 cm-1 = 0.120 cm-1 + 1/di

-0.0547 cm-1 = 1/di

-18.3 cmhi/ho = - di/dohi /(4.00 cm) = - (-18.3 cm)/(8.30 cm)hi = - (4.00 cm) • (-18.3 cm)/(8.30 cm)8.81 cm

A 4.00-cm tall light bulb is placed a distance of 8.30 cm from a double convex lens having a focal length of 15.2 cm. Determine the image distance and the image size.1/f = 1/do + 1/di1/(15.2 cm) = 1/(8.30 cm) + 1/di

0.0658 cm-1 = 0.120 cm-1 + 1/di

-0.0547 cm-1 = 1/di

-18.3 cmhi/ho = - di/dohi /(4.00 cm) = - (-18.3 cm)/(8.30 cm)hi = - (4.00 cm) • (-18.3 cm)/(8.30 cm)8.81 cm

A 4.00-cm tall light bulb is placed a distance of 35.5 cm from a diverging lens having a focal length of -12.2 cm. Determine the image distance and the image size.1/f = 1/do + 1/di1/(-12.2 cm) = 1/(35.5 cm) + 1/di

-0.0820 cm-1 = 0.0282 cm-1 + 1/di

-0.110 cm-1 = 1/di

-9.08 cm

hi/ho = - di/dohi /(4.00 cm) = - (-9.08 cm)/(35.5 cm)hi = - (4.00 cm) * (-9.08 cm)/(35.5 cm)1.02 cm

A 4.00-cm tall light bulb is placed a distance of 35.5 cm from a diverging lens having a focal length of -12.2 cm. Determine the image distance and the image size.1/f = 1/do + 1/di1/(-12.2 cm) = 1/(35.5 cm) + 1/di

-0.0820 cm-1 = 0.0282 cm-1 + 1/di

-0.110 cm-1 = 1/di

-9.08 cm

hi/ho = - di/dohi /(4.00 cm) = - (-9.08 cm)/(35.5 cm)hi = - (4.00 cm) * (-9.08 cm)/(35.5 cm)1.02 cm

Determine the image distance and image height for a 5-cm tall object placed 45.0 cm from a double convex lens having a focal length of 15.0 cm.

Given: f =15 cm and do = 45 cm and ho= 5 cmUse 1 / f = 1 / do + 1 / di to solve for di

1 / (15 cm) = 1 / (45 cm) + 1 / di1 / (15 cm) - 1 / (45 cm) = 1 / di

0.0444 cm-1 = 1 / di

di = 1 / (0.0444 cm-1)di = 22.5 cm

Then use hi / ho = - di / doto solve for hi

hi / (5 cm) = - (22.5 cm) / (45 cm) hi = - (5 cm) • (22.5 cm) / (45 cm)hi = -2.5 cm

Determine the image distance and image height for a 5-cm tall object placed 45.0 cm from a double convex lens having a focal length of 15.0 cm.

Given: f =15 cm and do = 45 cm and ho= 5 cmUse 1 / f = 1 / do + 1 / di to solve for di

1 / (15 cm) = 1 / (45 cm) + 1 / di1 / (15 cm) - 1 / (45 cm) = 1 / di

0.0444 cm-1 = 1 / di

di = 1 / (0.0444 cm-1)di = 22.5 cm

Then use hi / ho = - di / doto solve for hi

hi / (5 cm) = - (22.5 cm) / (45 cm) hi = - (5 cm) • (22.5 cm) / (45 cm)hi = -2.5 cm

Determine the image distance and image height for a 5-cm tall object placed 30.0 cm from a double convex lens having a focal length of 15.0 cm.

Given: f =15 cm and do = 30 cm and ho = 5 cmUse 1 / f = 1 / do + 1 / di

1 / (15 cm) = 1 / (30 cm) + 1 / di1 / (15 cm) - 1 / (30 cm) = 1 / di

0.0.333 cm-1 = 1 / di

di = 1 / (0.0333 cm-1)di = 30 cmThen use hi / ho = -di / do

hi / (5 cm) = - (30 cm) / (30 cm) hi = - (5 cm) • (30 cm) / (30 cm)hi = -5.0 cm

Determine the image distance and image height for a 5-cm tall object placed 30.0 cm from a double convex lens having a focal length of 15.0 cm.

Given: f =15 cm and do = 30 cm and ho = 5 cmUse 1 / f = 1 / do + 1 / di

1 / (15 cm) = 1 / (30 cm) + 1 / di1 / (15 cm) - 1 / (30 cm) = 1 / di

0.0.333 cm-1 = 1 / di

di = 1 / (0.0333 cm-1)di = 30 cmThen use hi / ho = -di / do

hi / (5 cm) = - (30 cm) / (30 cm) hi = - (5 cm) • (30 cm) / (30 cm)hi = -5.0 cm

Determine the image distance and image height for a 5-cm tall object placed 20.0 cm from a double convex lens having a focal length of 15.0 cm.

Given: f = 15 cm and do = 20 cm and ho = 5 cmUse 1 / f = 1 / do + 1 / di

1 / (15 cm) = 1 / (20 cm) + 1 / di1 / (15 cm) - 1 / (20 cm) = 1 / di

0.0167 cm-1 = 1 / di

di = 1 / (0.0167 cm-1)di = 60.0 cmThen use hi / ho = -di / do

hi / (5 cm) = - (60 cm) / (20 cm) hi = - (5 cm) • (60 cm) / (20 cm)hi = -15.0 cm

Determine the image distance and image height for a 5-cm tall object placed 20.0 cm from a double convex lens having a focal length of 15.0 cm.

Given: f = 15 cm and do = 20 cm and ho = 5 cmUse 1 / f = 1 / do + 1 / di

1 / (15 cm) = 1 / (20 cm) + 1 / di1 / (15 cm) - 1 / (20 cm) = 1 / di

0.0167 cm-1 = 1 / di

di = 1 / (0.0167 cm-1)di = 60.0 cmThen use hi / ho = -di / do

hi / (5 cm) = - (60 cm) / (20 cm) hi = - (5 cm) • (60 cm) / (20 cm)hi = -15.0 cm

Determine the image distance and image height for a 5-cm tall object placed 10.0 cm from a double convex lens having a focal length of 15.0 cm.

Given: f = 15 cm and do = 10.0 cm and ho = 5 cmUse 1 / f = 1 / do + 1 / di

1 / (15 cm) = 1 / (10 cm) + 1 / di1 / (15 cm) - 1 / (10 cm) = 1 / di

-0.0333 cm-1 = 1 / di

di = 1 / (-0.0333 cm-1)di = - 30.0 cmThen use hi / ho = - di / do

hi / (5 cm) = - (-30 cm) / (10 cm) hi = - (5 cm) • (-30 cm) / (10 cm)hi = 15.0 cm

Determine the image distance and image height for a 5-cm tall object placed 10.0 cm from a double convex lens having a focal length of 15.0 cm.

Given: f = 15 cm and do = 10.0 cm and ho = 5 cmUse 1 / f = 1 / do + 1 / di

1 / (15 cm) = 1 / (10 cm) + 1 / di1 / (15 cm) - 1 / (10 cm) = 1 / di

-0.0333 cm-1 = 1 / di

di = 1 / (-0.0333 cm-1)di = - 30.0 cmThen use hi / ho = - di / do

hi / (5 cm) = - (-30 cm) / (10 cm) hi = - (5 cm) • (-30 cm) / (10 cm)hi = 15.0 cm

A magnified, inverted image is located a distance of 32.0 cm from a double convex lens with a focal length of 12.0 cm. Determine the object distance and tell whether the image is real or virtual.

Given: f = 12 cm and di = + 32 cm (inverted images are real and have + image distances)Use 1 / f = 1 / do + 1 / di

1 / (12 cm) = 1 / do + 1 / (32 cm) 1 / (12 cm) - 1 / (32 cm) = 1 / do

0.052083 cm-1 = 1 / do

do = 1 / (0.052083 cm-1)do = 19.2 cm

A magnified, inverted image is located a distance of 32.0 cm from a double convex lens with a focal length of 12.0 cm. Determine the object distance and tell whether the image is real or virtual.

Given: f = 12 cm and di = + 32 cm (inverted images are real and have + image distances)Use 1 / f = 1 / do + 1 / di

1 / (12 cm) = 1 / do + 1 / (32 cm) 1 / (12 cm) - 1 / (32 cm) = 1 / do

0.052083 cm-1 = 1 / do

do = 1 / (0.052083 cm-1)do = 19.2 cm

ZINGER: An inverted image is magnified by 2 when the object is placed 22 cm in front of a double convex lens. Determine the image distance and the focal length of the lens.

Given: do = 22 cm and M = -2 (inverted images have negative image heights and therefore negative magnification values)The magnification of an image is both the hi / ho ratio and the -di / do ratio. Setting the -di / do ratio equal to -2 allows one to determine the image distance:M = -di / do = -2 -di / (22 cm) = -2di = 2 • (22 cm)di = 44 cmNow substitute the di and do values into the lens equation 1 / f = 1 / do + 1 / di to solve for the focal length.1 / f = 1 / (22 cm) + 1 / (44 cm) 1 / f = 0.06818 cm-1

f = 1 / (0.06818 cm-1)f = 14.7 cm

ZINGER: An inverted image is magnified by 2 when the object is placed 22 cm in front of a double convex lens. Determine the image distance and the focal length of the lens.

Given: do = 22 cm and M = -2 (inverted images have negative image heights and therefore negative magnification values)The magnification of an image is both the hi / ho ratio and the -di / do ratio. Setting the -di / do ratio equal to -2 allows one to determine the image distance:M = -di / do = -2 -di / (22 cm) = -2di = 2 • (22 cm)di = 44 cmNow substitute the di and do values into the lens equation 1 / f = 1 / do + 1 / di to solve for the focal length.1 / f = 1 / (22 cm) + 1 / (44 cm) 1 / f = 0.06818 cm-1

f = 1 / (0.06818 cm-1)f = 14.7 cm