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Light, Vision and Optics

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Reflection and Refraction

Light encountering the interface between media may be reflected and/or refracted.

Reflection is bouncing off a barrier or interface. Refraction is bending of light's path at the interface between two media. All angles are to be measured from the normal to the surface. The incident angle and the reflected angle are equal. The refracted angle may be larger or smaller than the incident angle.
The index of refraction, n, is a ratio that compares the speed of light in a vacuum to the speed of light in another medium.
n = c/v where c = speed of light in a vacuum = 3x108 m/s
The speed of light in all other media is less than in a vacuum. Therefore, n is always larger than one. It may be thought of as the measure of the slowness of light in a medium: the slower light travels in medium, the larger n will be for that medium. Since n divides speed by speed it has no units.
The frequency of light does not change as it enters a different medium, but its wavelength does. Since
v = fL where f = frequency and L = wavelength
then
v1/v2 = L1/L2

EXAMPLE

1.
a. What is the speed of light in water, if water's index of refraction is 1.33?
b. If the wavelength of a ray of light in a vacuum is 1x10-6 m, what is its wavelength in water?

ANSWER

1.
a. n = c/v
1.33 = (3 X 108 m/s)/v
v = 2.26x108 m/s
b. v1/v2 = L1/L2
(2.26x108 m/s)/(3x108 m/s) = L1/(1x10-6 m)
L1 = 7.52x10-7 m

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SNELL'S LAW

Snell's Law describes the relationship between the incident and refracted rays of light.

Snell's Law relates the index of refraction of a medium to the angle of light from the normal at the interface. Notice that it does not matter which ray is incident or refracted. (related search term: snell)
n1sinø1 = n2sinø2
Notice that to maintain the equality, as the index of refraction of a medium increases, ø must decrease. In other words, in a slower medium, the light will be refracted more closely to the normal.

Example Problem

2. Light traveling through air (n = 1.003) hits the surface of water (n = 1.33) with an incident angle of 30º. What is the refracted angle?

ANSWER

2. n1sinø1 = n2sinø2
(1.003)sin30º = (1.33)sinø2
ø2 = 22.2º

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CRITICAL ANGLE AND TOTAL INTERNAL REFLECTION

As light travels from a slow medium to a faster medium, such as from water into air, it is refracted away from the normal.As light travels from a slow medium to a faster medium, such as from water into air, it is refracted away from the normal. As the incident angle increases, the refracted angle increases.

At a large enough incident angle (critical angle), the refracted angle is 90 degrees and the refracted ray travels along the interface between the two media.At a large enough incident angle ("critical angle"), the refracted angle is 90º and the refracted ray travels along the interface between the two media.
nisinøc = nrsin90º
sinøc = nr/ni

If no light travels out of the slower medium; it is all reflected internally.At angles larger than øc no light travels out of the slower medium; it is all reflected internally.

Example Problem

3. What is the critical angle of light traveling from water into air?

ANSWER

3.

nisinøc = nrsin90º
(1.33)sinøc = (1.003)sin90º
sinøc = (1.003)/(1.33)
øc = 48.9º

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MIRROR EQUATION

Example Problem

4. A concave mirror forms a real image at 14 cm from the mirror surface along the principal axis. If the corresponding object is at a 29 cm distance, what is the mirror's focal length?

ANSWER

Apply the Mirror Equation, 1/do + 1/di = 1/f
1/ 29 + 1/14 = 1/f
f = 9.4 cm

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Light, Vision and Optics Problems

(See bottom of page for answers.)

A. Refraction or Dispersion Problems

(See bottom of page for answers.)

A ray of light traveling through the air enters a layer of glass, then a layer of water.1. Sketch a ray of light traveling through the air as it enters a layer of glass, then a layer of water, and then into the air again. The index of refraction of glass > the index of refraction of water > the index of refraction of air.

2. A beam of light passes through air, into a plate of flint glass (n = 1.66), then a plate of transparent acrylic (n = 1.50), then into the air again. If the light strikes the flint glass with an incident angle ø1 and emerges at an angle of 40º (in air again) with respect to the normal, calculate the value of ø1 . Draw a ray diagram for the light path.

3. Why does a spear fisherman aim at a fish's tail?

4. Why is yellow light used in fogs?

5. Light travels through a vacuum at about 300 million meters per second and through water at about 226 million meters per second.  Calculate the index of refraction for water. Can the absolute index of refraction be less than one?

6. A plate of glass (refractive index of n = 1.8) is placed on a photograph. The photograph appears to be 5 mm closer to the observer than without the glass. How thick is the glass?

7. An image of a distant object is found to be located on the back surface of a glass sphere. What is the refractive index of the glass?

8. An optic lens is made of flint glass (n = 1.66) with a coating of magnesium fluoride (n = 1.38). What thickness of coating will minimize reflected light of wavelength 543 nm?

B. Lens Problems

(See bottom of page for answers.)

1. The left side of a biconvex lens has radius of curvature of 20 cm. The right side of the lens is shaped so that an object located 36 cm to the left of the lens will form a real image 72 cm to the right of the lens. What is the radius of the curvature of the right side if the lens has an index of refraction of 1.5?

2. A converging lens of focal length 20 cm is 50 cm from a converging lens of focal length 5 cm.
a. Find the position of the image of an object placed 40 cm in front of the first lens.
b. If height of object is 2 cm, what is height of the final image? is it real or virtual?

3. How far from a 50-mm focal length lens must an object be to form a real image magnified by a factor of three?

4. An object is located 1.25 m in front a screen. Determine the focal length of a lens that forms a real inverted image on the screen 4 times the height of the object.

C. Reflection and Mirror Problems

(See bottom of page for answers.)

1. A convex mirror with radius of curvature of 10 cm produces a virtual image one third the size of the object. a. Where is the object located?
b. Draw an accurate ray diagram.

2. The mirrors on some cars warn that objects may be closer than they appear.
a. What kind of mirror is being used?
b. Why is the advantage of that type of mirror?

3. When an object is placed 20.4 cm in front of a particular convex mirror, the image is found 5.95 cm behind the mirror. When a second object with twice the height of the first one is placed in front of the mirror, its image has the same height as the first image. Where is the second object located?

5. An object 25 cm in front of a concave mirror produces an upright image magnified by a factor of two. What is the radius of curvature of the mirror?

A man looks at his reflection in a mirror.

6. A man stands 3 feet in front of a vertical mirror.  His eyes are 5’8” above the ground.  The top of the mirror is level with his eyes.  How long does the mirror have to be so that he can see his shoes?  Does his distance away from the mirror matter?

D. Other Problems

(See bottom of page for answers.)

1. What is the illuminance on a surface 3 meters from 72-candela light source?

2. if you were driving towards the sun late in the afternoon how can you reduce the glare from the road?

3. Mars is 5.6x1010 m from Earth at the point of closest approach. Determine the minimum time required for a person's voice to reach Earth from Mars.

4. A first-order maximum is observed 0.350mm from the central maximum on a distant screen when light of 600nm wavelength is diffracted. What wavelength of light produces a third-order maximum 0.870 mm from the central maximum using the same diffraction grating?

5.
a. What color would appear on a screen illuminated simultaneously by a red light source and a green light source?
b. What color would a blue object appear to be when illuminated simultaneously by a red light source and a green light source?

6. If the intensity of light is 7.2 W/m2 at a distance of 3.6 meters, find its intensity at 1.2 meters.

7. Unpolarized light travels through two polarizers: the second at a 30º angle to the first. If the original intensity of light is 4 W/m2, what is the intensity of light after the second polarizer?

8. In a single-slit Fraunhofer diffraction pattern, which fringe number has a peak intensity that is 0.5 percent of the central fringe intensity?

9. Calculate the light intensity 1.45 m from a 100.0 W light bulb.

Answers to Light, Vision and Optics Problems

Selected solutions are printed below.
For solutions to all the problems on this page click here.

When looking through the plate, the photograph appears to be 5 mm closer to the observer than without it.

A6.

OT is the real depth of the photograph.
PT is the apparent depth of the photograph.
Now the diagram appears 5 mm closer. Thus OP =5mm = 0.5 cm
Now, we know that,
Refractive index n = Real depth/Apparent depth
Thus,
n = OT/PT
1.8 = OT/(OT – OP) [ as, n =1.8, PT = OT – OP]
1.8 = OT/(OT – 0.5)
1.8OT – 0.9 = OT
0.8OT = 0.9
OT = 9/8
Thus the thickness of the plate is 9/8 cm.

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B4.

Givens: m = -4
di + do = 1.25 m

Solution:

Since m = -( di / do )
-4 = -( di / do )
di = 4do

Since di + do = 1.25 m,
4do+ do = 1.25 m
do =0 .25 m

Therefore,
di = 1.00 m

Using the lens equation: (1/f ) = ( 1/di ) + (1/do )
(1/f ) = ( 1/1.00 ) + (1/0.25 ) [Object distance is positive since it is on the side from which light is coming; image distance is positive since it is on the opposite side of the lens from the object]

f = 0.20 m

Since the focal length is positive, the lens is convex.

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C4.

In dark conditions, the pupils of the eye open wide to allow more light in.  At the same time, they also allow more reflected light out.  This reflected light is not noticeable in bright daylight conditions, but is very noticeable in the dark.  “Red eye” is a common problem during flash photography as the human retina is red in color because of its blood vessels.  Some cameras have a red-eye setting that causes the flash to go off several times before the picture is taken. The earlier flashes close the pupils to minimize reflection from the retina.

Animal and human eyes are constructed differently.  Animals have a reflective layer behind the retina that provides more light to the retina. This layer is why animals see better in the dark, and also determines the color of the light reflected from their eyes.

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D6.

Intensity is inversely proportional to the square of the separation:
I2 / I1 = d12 / d22
I2 /(7.2) = (3.6) 2 / (1.2) 2
>I2 = 64.8 W/m2

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