OPTICSOPTICS

I.I. IMAGESIMAGES

A. Definition- A. Definition- An image is formed An image is formed where light rays originating from where light rays originating from the same point on an object the same point on an object intersect on a surface or appear intersect on a surface or appear to intersect for the observerto intersect for the observer

B. B. Real Image- formed when light Real Image- formed when light rays from a common point pass rays from a common point pass though an optical system that though an optical system that causes them to converge and causes them to converge and intersect at a pointintersect at a point

1. A real image can be 1. A real image can be projectedprojected on a screen when placed where the on a screen when placed where the image is formed.image is formed.

a. Lenses in a slide projector or a a. Lenses in a slide projector or a camera produce real imagescamera produce real images

C. Virtual C. Virtual Image-formed when the Image-formed when the light rays from a common point light rays from a common point pass through or are reflected by an pass through or are reflected by an optical system that causes them to optical system that causes them to diverge and diverge and appearappear to come to a to come to a single point. single point.

1. cannot be 1. cannot be projectedprojected on a screen on a screen because no light from the object because no light from the object actually reaches the point where the actually reaches the point where the image image appearsappears to be located to be located

a. Example Reflection in a plane a. Example Reflection in a plane mirror. The rays reaching the mirror. The rays reaching the observers eyes actually come from the observers eyes actually come from the object. They are reflected by the object. They are reflected by the mirror in such a way that they mirror in such a way that they appear appear to come from the imageto come from the image

EACH RAY IS REFLECTED IN A REGULAR WAY ( MEANING THAT THE ANGLEOF REFLECTION EQUALS THE ANGLE OF INCIDENCE). IF THE REFLECTEDRAY IS PROJECTED BACK THEY FORM THE LOCATION OF THE IMAGE BEHIND THE MIRROR, A VIRTUAL IMAGE

II. CONSTRUCTING A RAY DIAGRAM FOR A II. CONSTRUCTING A RAY DIAGRAM FOR A PLANE MIRROR.PLANE MIRROR.

1.1. DRAW A RAY FROM THE OBJECT TO THE DRAW A RAY FROM THE OBJECT TO THE MIRROR.MIRROR.

2.2. THEN CONSTRUCT THE REFLECTED RAY.THEN CONSTRUCT THE REFLECTED RAY.

3.3. EXTEND THE REFLECTED RAY BEHIND EXTEND THE REFLECTED RAY BEHIND THE MIRROR. THE MIRROR. Use dotted linesUse dotted lines

4.4. REPEAT FOR ONE MORE RAY.REPEAT FOR ONE MORE RAY.

5.5. WHERE THE RAYS INTERSCET BEHIND WHERE THE RAYS INTERSCET BEHIND THE MIRROR IS WHERE IS IMAGE IS THE MIRROR IS WHERE IS IMAGE IS LOCATED.LOCATED.

IMAGES PRODUCED BY A PLANE MIRROR ARE VIRTURAL, THE SAME SIZE AS THE OBJECT ANDLOCATED AN EQUAL DISTANCE BEHIND THE MIRROR AS THE OBJECT WAS IN FRONT OF THE MIRROR

► III. Images formed by spherical III. Images formed by spherical mirrorsmirrors

A. DefinitionsA. Definitions

1. Concave 1. Concave mirrors-Generally mirrors-Generally form real imagesform real images

2. Convex Mirror-Always form virtual 2. Convex Mirror-Always form virtual imagesimages

3. Convex Lens-Generally form real 3. Convex Lens-Generally form real imagesimages

4. Concave Lens-Always form 4. Concave Lens-Always form virtual imagesvirtual images

► Constructing Ray Diagrams for MirrorsConstructing Ray Diagrams for Mirrors► REMEMBER MIRRORS REMEMBER MIRRORS REFLECT LIGHTREFLECT LIGHT

1.1. Concave mirrorsConcave mirrors

a. Use the real focal point, located in a. Use the real focal point, located in front of the mirror front of the mirror

b. A real image is always formed b. A real image is always formed except except when the object is located between F when the object is located between F and and

the mirror.the mirror.

c. If the rays do not intersect in front the c. If the rays do not intersect in front the mirror extend them behind the mirror to mirror extend them behind the mirror to locate the virtual imagelocate the virtual image

2.2. Convex MirrorsConvex Mirrors

a. A virtual image is always formeda. A virtual image is always formed

b. Use the virtual focal point (located b. Use the virtual focal point (located behind the mirror) behind the mirror)

c. The rays do not intersect in front c. The rays do not intersect in front the mirror, extend them behind the the mirror, extend them behind the mirror to locate the virtual imagemirror to locate the virtual image

Lenses- Refract lightLenses- Refract light

Constructing Ray Diagrams for LensesConstructing Ray Diagrams for Lenses► Convex LensesConvex Lenses

a. Use the real focal point located on the a. Use the real focal point located on the

side opposite of the object.side opposite of the object.

b. Will always form real images except b. Will always form real images except

when the object is located between F when the object is located between F

and the lensand the lens

c. If the rays do not intersect behind of the c. If the rays do not intersect behind of the lens then extend them to the front of the lens then extend them to the front of the lens to locate the virtual imagelens to locate the virtual image

2.2. Concave LensConcave Lens

a. Always form virtual imagesa. Always form virtual images

b. Use the virtual focal point (located b. Use the virtual focal point (located on the same side of the lens as the on the same side of the lens as the object ) object )

c. The rays will not intersect behind c. The rays will not intersect behind the lens so extend them to in front of the lens so extend them to in front of the lens to locate the virtual imagethe lens to locate the virtual image

►VIVI Mirror/lens EquationsMirror/lens Equations► A. Mathematically relates the locations and A. Mathematically relates the locations and

size of the image and the objectsize of the image and the object

B. B. 11 + + 1 1 = = 11

ddo o ddii f f

WhereWhere

ddo o = = distance from object to mirror/lensdistance from object to mirror/lens

dd11 = = distance from image to mirror/lensdistance from image to mirror/lens

f = f = focal lengthfocal length

Units- any unit of length all must be the sameUnits- any unit of length all must be the sameFor convex mirror and concave lens make f

negative

C. C. SSoo = = ddoo

SSii d dii

WhereWhere

SSoo = = size of objectsize of object

SSii = = size of imagesize of image

ddoo = = distance from object to distance from object to mirror/lensmirror/lens

ddii = = distance from image to mirror/lensdistance from image to mirror/lens

Units- any units of length, all the sameUnits- any units of length, all the same

Example:Example:

An object 2.0 cm high is 30.0 cm from a An object 2.0 cm high is 30.0 cm from a concave mirror. The focal length of concave mirror. The focal length of the mirror is 10.0cm. A. What is the the mirror is 10.0cm. A. What is the location of the image? B. What is the location of the image? B. What is the size of the image?size of the image?

11 + + 1 1 = = 1 1 11 + + 11 = = 11

ddo o ddii f 30cm d f 30cm dii 10.0cm 10.0cm

11 = 0.06667 d = 0.06667 dii= 15 cm= 15 cm

ddii

►B. B. SSoo = = ddo o 2.0cm = 30.0cm

SSii d dii S Sii 15cm 15cm

Cross multiplyCross multiply

2X 15 = 30 S2X 15 = 30 Sii

SSii = 1cm = 1cm

Example 2: Find the location and size of Example 2: Find the location and size of the image of a 2.0cm high object the image of a 2.0cm high object located 5.0cm in front of a convex lens located 5.0cm in front of a convex lens of focal length 10.0cm.of focal length 10.0cm.

11 + + 1 1 = = 1 1 1 1 + + 1 1 = = 1 1

dd o o ddii f 5.0cm d f 5.0cm dii 10.0cm 10.0cm

11 = -.10 d = -.10 dii= -10.0cm negative = -10.0cm negative indicatesindicates

ddi i virtual imagevirtual image

SSoo = = ddoo 2.0cm2.0cm = = 5.0cm5.0cm

SSii d dii S Sii -10.0cm -10.0cm

Cross multiplyCross multiply

-20 = 5 S-20 = 5 Sii

SSii = -4 cm = -4 cm

Negative means virtual imageNegative means virtual image

Example 3: calculate the position of the Example 3: calculate the position of the image of an object located 15cm in image of an object located 15cm in front of a convex mirror of focal length front of a convex mirror of focal length -10.ocm.-10.ocm.

11 + + 1 1 = = 11 1 1 + + 1 1 = = 11

ddo o ddii f 15cm d f 15cm dii - -10.0cm10.0cm

11= -0.1667 d= -0.1667 dii =-6.0cm negative =-6.0cm negative meansmeans

ddi i a virtual imagea virtual image

VIII.VIII. Lens DefectsLens Defects

A. Two TypesA. Two Types

1. Chromatic Aberration1. Chromatic Aberration

a. a. Occurs because different colors of Occurs because different colors of light do not focus at the same point.light do not focus at the same point.

b. Lens edges act like a prism, different b. Lens edges act like a prism, different wavelengths bent at slightly different wavelengths bent at slightly different angles. angles.

c. Often occurs in camerasc. Often occurs in cameras

d. Reduced by using combination of d. Reduced by using combination of lenses made of lenses made of different different types of glass. types of glass.

2. Spherical Aberration2. Spherical Aberration

a. a. Occurs because the Occurs because the spherical shape of the lens is not spherical shape of the lens is not ideal for converging light to a ideal for converging light to a single point.single point.

b. Can be reduced by restricting b. Can be reduced by restricting the light beam so it is incident close to the light beam so it is incident close to the center of the lens.the center of the lens.

c. accomplished by c. accomplished by reducingreducing the the size of the lens opening. (the aperture) size of the lens opening. (the aperture)

►DIFFRACTION AND DIFFRACTION AND INTERFERENCE OF LIGHTINTERFERENCE OF LIGHT

A. Diffraction- A. Diffraction- Bending of waves around Bending of waves around a boundary. Readily seen with a boundary. Readily seen with water waves.water waves.

B. Young’s Double Slit experimentB. Young’s Double Slit experiment

1. Young showed that light bends or 1. Young showed that light bends or diffracts around boundaries.diffracts around boundaries.

2. Light was allowed to fall on two 2. Light was allowed to fall on two closely spaced narrow slits.closely spaced narrow slits.

3. The light passing through each slit 3. The light passing through each slit was diffracted.was diffracted.

4. The spreading light from the two 4. The spreading light from the two slits overlap. (Interfere)slits overlap. (Interfere)

5. When the light from the two slits fell on a 5. When the light from the two slits fell on a screen a pattern of light and dark bands screen a pattern of light and dark bands were observed. Called were observed. Called interference interference fringesfringes..

6.Bright bands correspond to 6.Bright bands correspond to constructive interference.constructive interference. (two (two crests or two troughs)crests or two troughs)

7. Dark bands correspond to 7. Dark bands correspond to destructive destructive interference. interference. (a crest and a trough)(a crest and a trough)

8. 8. Central Band- Central Band- zero order bandzero order band

9. bright bands on either side of the central 9. bright bands on either side of the central band are referred to as the band are referred to as the first order first order band, second- order bandband, second- order band and so on. and so on.

► J. Mathematical RelationshipJ. Mathematical Relationship

X = X = λλLL

dd

Where Where X is the distance between X is the distance between any two bright bandsany two bright bands

..λλ = = wavelength of light usedwavelength of light used

L= L= distance from the slits to the distance from the slits to the screenscreen..

d.= d.= distance between the slits.distance between the slits.

Units of length all must be the sameUnits of length all must be the same

1.1. Example Monochromatic light is Example Monochromatic light is incident on a pair of slits 1.95 X 10incident on a pair of slits 1.95 X 10-5-5 m apart. The distance between the m apart. The distance between the first two bright bands is 2.11 X 10first two bright bands is 2.11 X 10-2-2m. m. If the distance between the slits and If the distance between the slits and the screen is 0.600m.the screen is 0.600m.

A) calculate the wavelengthA) calculate the wavelength

B) state the color of the lightB) state the color of the light

d= 1.95 X 10d= 1.95 X 10-5-5 m X= 2.11 X 10 m X= 2.11 X 10-2-2mm

L = 0.600mL = 0.600m

X = X = λλLL

dd

2.11 X 102.11 X 10-2-2m = m = λλ((0.600m)0.600m)

1.95 X 101.95 X 10-5-5 m m

λλ = = 2.11 X 102.11 X 10-2-2m (1.95 X 10m (1.95 X 10-5-5 m) m)

0.60.600m00m

λλ = 6.86 X 10 = 6.86 X 10-7-7mm

This is red lightThis is red light

III.III. Single Slit DiffractionSingle Slit Diffraction

A. Diffraction pattern produced A. Diffraction pattern produced differs from that of a double slit.differs from that of a double slit.

1. The central bright band is 1. The central bright band is much much widerwider than any of the other than any of the other bright bandsbright bands

2. The intensity of the central 2. The intensity of the central band is band is greatergreater than the intensity than the intensity of any of the other bright bands.of any of the other bright bands.

3. Mathematical Relationship3. Mathematical Relationship

X= X= λλLL

ww

Where Where

X = X = separation distance between separation distance between central band and dark bandcentral band and dark band

..λλ = wavelength of light used= wavelength of light used

L = L = distance from the slit to the distance from the slit to the screenscreen..

W = W = width of the slitwidth of the slit..

Units of length all must be the sameUnits of length all must be the same

4.4. Example: Monochromatic orange Example: Monochromatic orange light of wavelength 605nm falls on a light of wavelength 605nm falls on a single slit of width 0.095mm. The silt single slit of width 0.095mm. The silt is located 85cm from a screen. How is located 85cm from a screen. How far is the first dark band?far is the first dark band?

λλ= 605nm = 605 X 10= 605nm = 605 X 10-9-9mm

w = 0.095mm = .000095mw = 0.095mm = .000095m

L = 85cm= .85mL = 85cm= .85m

X= X= λλLL

ww

X=X=605 X 10605 X 10-9-9m (.85m)m (.85m)

.000095m.000095m

X = .0054m= 5.4mmX = .0054m= 5.4mm

IV.IV. Diffraction GratingsDiffraction Gratings

A. Used in practice to measure the A. Used in practice to measure the wavelength of light.wavelength of light.

B. Made by scratching very fine lines B. Made by scratching very fine lines with a diamond point on glass.with a diamond point on glass.

1. The clear lines on the glass 1. The clear lines on the glass serve as the slits.serve as the slits.

2. There may be as many as 2. There may be as many as 10,000 lines per cm.10,000 lines per cm.

C. Interference Pattern similar to C. Interference Pattern similar to double silt except:double silt except:

1. The bright bands are 1. The bright bands are narrowernarrower and the dark bands are and the dark bands are broader.broader.

2. gives a more precise 2. gives a more precise measurement than measurement than double slit.double slit.

D.D. Mathematical relationshipMathematical relationship

X = X = λλLL

dd

Examples:Examples:

A good diffraction grating has 2.50 X A good diffraction grating has 2.50 X 101033lines/cm. What is the distance lines/cm. What is the distance between two lines in the grating?between two lines in the grating?

1/2.50 X 101/2.50 X 1033lines/cm= .00040cmlines/cm= .00040cm

Example: Using a grating with a spacing Example: Using a grating with a spacing of 4.00 X 10of 4.00 X 10-4-4cm, a red line appears 16.5 cm, a red line appears 16.5 cm from the central line on a screen. cm from the central line on a screen. The screen is 1.00m from the grating. The screen is 1.00m from the grating. What is the wavelength of the red light.What is the wavelength of the red light.

X = X = λλLL

dd

16.5cm = 16.5cm = λλ(100cm)(100cm)

4.00 X 104.00 X 10-4-4cmcm

λλ=6.6 X 10=6.6 X 10-5-5 cm cm