Question 1

A glass plate whose index of refraction is 1.56 is immersed in a liquid. The surface of the glass is inclined at an angle of 42.0° with the vertical. A horizontal ray in the glass is incident on the interface. When the liquid is a certain alcohol, the incident ray arrives at the interface at the critical angle. What is the index of refraction of the alcohol?

Accepted Answer

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Question 2

The image of a plant is 4.0 cm from a concave spherical mirror having a radius of curvature of 10 cm. Where is the plant relative to the mirror?

Accepted Answer

The mirror equation, $ \frac{1}{f} = \frac{1}{v} + \frac{1}{u} $, where $f$ is the focal length, $v$ is the image distance, and $u$ is the object distance, can be used here. The focal length $f = \frac{R}{2} = 5 $ cm (since the radius of curvature $R = 10$ cm). Given $v = -4.0$ cm (negative because the image formed by concave mirrors is on the same side as the object), we can solve for $u$. Substituting the values, we get $ \frac{1}{5} = \frac{1}{-4} + \frac{1}{u} $, solving for $u$ gives $u \approx -2.2$ cm, indicating the object is 2.2 cm in front of the mirror (negative sign indicates the object is in front of the mirror).

Question 3

How far are you from your image when you stand 0.750 m in front of a vertical plane mirror?

Accepted Answer

The distance from you to your image in a plane mirror is twice the distance between you and the mirror, so 0.750 m x 2 = 1.50 m.

Question 4

A tank holds a 1.44-m thick layer of oil that floats on a 0.96-m thick layer of brine. Both liquids are clear and do not intermix. Point O is at the bottom of the tank, on a vertical axis. The indices of refraction of the oil and the brine are 1.40 and 1.52, respectively. A ray originating at O reaches the brine-oil interface at the critical angle. What is the distance of this point from the axis?

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Question 5

The spherical side mirror on a car is convex and has a radius of curvature of 25 cm. Another car is following, 20 m behind the mirror. If the height of the car is 1.6 m, how tall is its image?

Accepted Answer

The mirror equation for a convex mirror, where $f$ is the focal length, $u$ is the object distance, and $v$ is the image distance, is given by $\frac{1}{f} = \frac{1}{v} + \frac{1}{u}$. The radius of curvature $R = 25$ cm, so the focal length $f = \frac{R}{2} = 12.5$ cm (remembering that for a convex mirror, the focal length is positive). The object distance $u = -2000$ cm (negative because it's real and in front of the mirror). Plugging these into the mirror equation gives $\frac{1}{12.5} = \frac{1}{v} - \frac{1}{2000}$. Solving for $v$ gives the image distance. The magnification $m = \frac{v}{u}$, and the height of the image $h_i = m \times h_o$, where $h_o = 160$ cm is the height of the object (the other car). Following through with the calculations, you find that the height of the image is approximately 0.99 cm, making E the correct answer.

Question 6

A man's face is 30 cm in front of a concave spherical shaving mirror. If the image is erect and 1.5 times as large as his face, what is the radius of curvature of the mirror?

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Question 7

A fish appears to be 2.00 m below the surface of a pond when viewed almost directly above by a fisherman. What is the actual depth of the fish? (n_water = 1.33)

Accepted Answer

The apparent depth of the fish is given by:
Apparent depth = Actual depth / Refractive index of water = d/1.33

We are given the apparent depth as 2.00 m. Substituting these values, we get:
2.00 = d/1.33

Solving for d, we get:
d = 2.00 x 1.33 = 2.66 m

Therefore, the actual depth of the fish is 2.66 m.

Question 8

An object is 12 cm in front of a concave spherical mirror, and the image is 3.0 cm in front of the mirror. What is the focal length of the mirror?

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Question 9

A convex spherical mirror with a focal length of magnitude 25 cm has a 4.0-cm tall flower placed 100 cm in front of it. What is the height of the image of the flower?

Accepted Answer

The height of the image can be found using the mirror equation and the magnification equation. First, find the image distance ($v$) using the mirror equation $\frac{1}{f} = \frac{1}{v} + \frac{1}{u}$, where $f$ is the focal length and $u$ is the object distance. Since the mirror is convex, $f$ is positive, and $u$ is -100 cm (object distance is considered negative in mirror equations). Solving for $v$, we get a small positive value indicating the image is virtual and on the same side as the object. Then, use the magnification equation $m = \frac{v}{u} = \frac{h_i}{h_o}$, where $h_i$ is the image height and $h_o$ is the object height. The magnification ($m$) will be less than 1 since $|v| < |u|$, indicating the image is smaller than the object. Plugging in the values and solving for $h_i$ gives a height of 0.80 cm, which matches choice A.

Question 10

A swimming pool is filled to a depth of 2.0 m. How deep does the pool appear to be from above the water, which has an index of refraction of 1.33?

Accepted Answer

The apparent depth of the pool is calculated using the formula $d' = \frac{d}{n}$, where $d'$ is the apparent depth, $d$ is the real depth (2.0 m), and $n$ is the index of refraction of water (1.33). Plugging in the values, we get $d' = \frac{2.0}{1.33} \approx 1.5$ m.

Question 11

A concave spherical mirror with a radius of 20 cm creates a real image 30 cm from the mirror. How far is the object from the mirror?

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Question 12

A convex spherical mirror with a radius of 50 cm has a 4.0 cm tall object placed 100 cm in front of it. What is the position of the image relative to the mirror?

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Question 13

An object 4.0 cm in height is placed 8.0 cm in front of a concave spherical mirror with a focal length of 10.0 cm. What is the position of its image in relation to the mirror, and what are the characteristics of the image?

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Question 14

A goldfish bowl is spherical, 8.0 cm in radius. A goldfish is swimming 3.0 cm from the wall of the bowl. Where does the fish appear to be to an observer outside? The index of refraction of water is 1.33. Neglect the effect of the glass wall of the bowl.

Accepted Answer

Using Snell's law of refraction,

n1*sinθ1 = n2*sinθ2

where n1 is the refractive index of air (approx. 1), θ1 is the angle of incidence (measured with respect to the normal), n2 is the refractive index of water (1.33), and θ2 is the angle of refraction.

Since the goldfish is in water, the angle of incidence is the angle between the line connecting the observer to the fish and the tangent to the surface of the water at the point where the fish is. Let O be the center of the sphere and T be the point where the line joining the observer to the fish intersects the surface of the water. Then, OT is perpendicular to the tangent at T. Since the angle at the center of a circle is twice the angle at the circumference subtended by the same arc, we have

θ1 = 2*OTC

where C is the point where the line joining the center of the sphere to the point of contact intersects the surface of the water. Similarly, the angle of refraction is the angle between the line connecting the observer to the fish and the tangent to the sphere (whose radius is r+3 cm) at the point where the fish appears to be. Let P be this point of apparent position. Then, the angle at the center of the sphere is

θ2 = 2*OPC'

where C' is the point where the line joining the center of the sphere to the point of apparent position intersects the surface of the water.

Since the angles at C, T, O, and P are all right angles, we have

OTC = C'OP = θ/2

where θ = angle TOP (angle of deviation).

Also, by symmetry, C' is the midpoint of OP. Therefore, by the Pythagorean theorem,

TC' = sqrt(r^2 - (r-3)^2) = sqrt(2*r*3-9)

Hence,

sin(θ/2) = TC'/r

Also,

sinθ1 = (r-3)/r

and

sinθ2 = (r-d)/r'

where d is the apparent distance of the fish from the wall (3 cm) and r' = r+n*d is the radius of the sphere of apparent position.

By Snell's law,

n1*sinθ1 = n2*sinθ2

or

sinθ2 = (n1/n2)*(r-3)/r

Therefore,

(r-d)/r' = (n1/n2)*(r-3)/r

or

d = (r/n2)*[(r-r')/(r'-r+3)]

Substituting the values, we get

d = (8.0 cm/1.33)*[(8.0 cm-(8.0 cm+3.0 cm))/(8.0 cm+3.0 cm-3.0 cm)]

= 1.498 cm

Therefore, the fish appears to be 3.0 cm - 1.498 cm = 1.502 cm inside the bowl from the observer's perspective. Rounding off to one decimal place, we get

d' = 1.5 cm

which is closest to option C (2.5 cm inside the bowl).

Question 15

An optical fiber made of glass with an index of refraction 1.50 is coated with a plastic with index of refraction 1.30. What is the critical angle of this fiber at the glass-plastic interface?

Accepted Answer

The critical angle can be found using the formula $	heta_c = \sin^{-1}(\frac{n_2}{n_1})$, where $n_1$ is the index of refraction of the material with higher index (in this case, glass) and $n_2$ is the index of refraction of the material with lower index (in this case, plastic). Plugging in the values, we get $	heta_c = \sin^{-1}(\frac{1.30}{1.50}) \approx 60.1°$. Therefore, the best choice is C.

Question 16

In the figure, the thin lens forms a real image of the object 94.0 cm from the object. What is the focal length of the lens?

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Question 17

The left-hand end of a glass rod is ground to a spherical surface. The glass has index of refraction 1.50. A small object 4.00 mm tall is placed on the axis of the rod, 37.0 cm to the left of the vertex of the spherical surface. The image is formed in the rod, 50.0 cm to the right of the vertex.
(a) What is the magnitude of the radius of curvature of the spherical surface at the end of the rod?
(b) What is the height of the image?

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Question 18

A ray of light traveling in air strikes the surface of a certain plastic slab at 63.0° with respect to the normal in air. It travels in the plastic slab at a 30.6° angle with respect to the normal. Find the critical angle for the plastic in air.

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Question 19

An object 3.4 mm tall is placed 25 cm from the vertex of a convex spherical mirror. The radius of curvature of the mirror has a magnitude of 52 cm.
(a) How far is the image from the vertex of the mirror?
(b) What is the height of the image?

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Question 20

A fisherman in a stream 39 cm deep looks downward into the water and sees a rock on the stream bed. How deep does the stream appear to the fisherman? The index of refraction of the water is 1.33.

Accepted Answer

When light travels from a medium of lower refractive index to a medium of higher refractive index, it bends towards the normal. And when light travels from a medium of higher refractive index to a medium of lower refractive index, it bends away from the normal. In this case, light is travelling from air (or vacuum) to water which has a higher refractive index. Therefore, the light will bend towards the normal.

Using Snell's Law, we can find the angle of incidence of the light at the water surface:

sin(i) = (refractive index of air/vacuum) / (refractive index of water) * sin(r)
sin(i) = (1/1.33) * sin(90) = 0.75
i = sin^-1(0.75) = 48.75 degrees

Now, we can use trigonometry to find the distance x, which is the depth of the rock below the surface of the water as seen by the fisherman:

sin(48.75) = (x/39)
x = 29 cm

Therefore, the stream appears to be 29 cm deep to the fisherman.

Quiz 34: Ray Optics