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LIGHT

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 = 3 X 10⁸ 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 1 X 10-6 m, what is its wavelength in water?

ANSWER TO EXAMPLE:

1.
a. n = c/v

1.33 = (3 X 10⁸ m/s)/v

v = 2.26 X 10⁸ m/s

b. v₁/v₂ = L₁/L₂

(2.26 X 10⁸ m/s)/(3 X 10⁸ m/s) = L1/(1 X 10^-6 m)

L₁ = 7.52 X 10^-7 m

SNELL'S LAW

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
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 TO EXAMPLE:

2. n₁sinø₁ = n₂sinø₂

(1.003)sin30º = (1.33)sinø₂

ø₂ = 22.2º

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 the incident angle increases, the refracted angle increases.

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.
n_isinø_c = n_rsin90º
sinø_c = n_r/n_i

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

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

ANSWER TO EXAMPLE:

3.

n_isinø_c = n_rsin90º

(1.33)sinø_c = (1.003)sin90º

sinø_c = (1.003)/(1.33)

ø_c = 48.9º

MIRROR EQUATION

EXAMPLE:
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 TO EXAMPLE:

Apply the Mirror Equation, 1/d_o + 1/d_i = 1/f

1/ 29 + 1/14 = 1/f

f = 9.4 cm

PROBLEMS:

A. Refraction or Dispersion Problems

1. A ray of light traveling through the air 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 thick light plate of flint glass(n = 1.66) rests on the top of a thick plate of transparent acrylic (n = 1.50). A beam of light is incident on top surface of the flint glass at an angle ø₁ (in air). the beam passes through the

glass and the acrylic 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 through the two plates of refracting material.

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.  It travels through water at about 226 million meters per second.  Calculate the index of refraction for water and explain why it could never be less than one.

6. A diagram is viewed through a plate of glass that is placed on the diagram and is known to have a refractive index of n = 1.8. When looking through the plate, the diagram appears to be 5 mm closer to the observer than without it. How thick is the plate?

7. A glass sphere is used to form an image of a distant object. The image is found to be located on the back surface of the sphere. What is the refractive index of the glass?

8. Find the minimum thickness of a layer of magnesium fluoride (n = 1.38) on flint glass (n = 1.66) that will cause destructive interference of reflected light of wavelength 543 nm near the middle of the visible spectrum.

B. Lens Problems

1. An object is located 36 cm to the left of biconvex lens of index of refraction 1.5. The left surface of the lens has radius of curvature of 20 cm. The right surface of the lens is to be shaped so that a real image will be formed at 72 cm to the right of the lens what is the required radius of the curvature of the second surface?

2. A converging lens of focal length 20 cm is separated by 50 cm from converging lens of focal length 5 cm.
a. Find the final 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, such as is used in many 35-mm cameras, must an object be positioned if it is to form a real image magnified in size by a factor of three?

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

C. Reflection and Mirror Problems

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 rear-view mirror on late model cars warns the user that objects may be closer than they appear.

a. What kind of mirror is being used.

b. Why was that type selected?

3. An object is located 20.4 cm in front of a convex mirror, the image being 5.95 cm behind the mirror. A second object, twice as tall as the first one, is placed in front of the mirror, but at a different location. The image of this second object has the same height as the other image. How far in front of the mirror is the second object located?

4. Why do cat's eyes shine at night?

5. A concave makeup mirror is designed so that a person 25 cm in front of it sees an upright image magnified by a factor of two. What is the radius of curvature of the 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

1. A 72-candela light source is located at a distance of 3 meters from a surface. What is the illuminance on the surface?

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

3. Someday a person will walk on Mars, which is 5.6x10¹⁰ m from Earth at the point of closest approach. Determine the minimum time that will be required for that person's voice to reach earth in seconds.

4. In a diffraction experiment, light of 600nm wavelength produces a first-order maximum 0.350mm from the central maximum on a distant screen. A second monochromatic source produces a third-order maximum 0.870 mm from the central maximum when it passes through the same diffraction grating. What is the wavelength of the light from the second source?

5.

a. A red light source and a green light source shine on a white screen. What color will the screen be?
b. If a blue object is placed in front of the screen, what color will it be? Explain your answer.

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

7. Originally, light is unpolarized. It then travels through 3 polarizers: the first with its transmission axis aligned vertical, the next at a 30º angle to the vertical and the last aligned at a 45º angle to the vertical (tilted the same direction as the first with respect to the vertical). 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 the intensity of the successive bright fringes falls off as we go out from the central maximum. Approximately, which fringe number has a peak intensity that is 0.5 percent of the central fringe intensity?

ANSWERS

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A6.

OT is the real depth of the diagram.
PT is the apparent depth of the diagram
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

d_i + d_o = 1.25 m

Solution:

Since m = -( d_i / d_o )

-4 = -( d_i / d_o )

d_i = 4d_o

Since d_i + d_o = 1.25 m,

4d_o+ d_o = 1.25 m

d_o =0 .25 m

Therefore,

d_i = 1.00 m

Using the lens equation: (1/f ) = ( 1/d_i ) + (1/d_o )

(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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