Question 1

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 2

A 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?

Accepted Answer

Using the mirror formula: 
1/f = 1/do + 1/di 
Where f is the focal length, do is the object distance and di is the image distance. 
Substituting: 
1/f = 1/12 + 1/-3 
Simplifying, we get: 
f = -2.4 cm 
Since the mirror is concave, the focal length is negative. Therefore, the answer is D) 2.4 cm.

Question 3

A ray of light consisting of blue light (wavelength 480 nm) and red light (wavelength 670 nm) is incident on a thick piece of glass at 80°. What is the angular separation between the refracted red and refracted blue beams while they are in the glass? (The respective indices of refraction for the blue light and the red light are 1.4636 and 1.4561.)

Accepted Answer

The angular separation can be calculated using Snell's law, which states $n_1 \sin(\theta_1) = n_2 \sin(\theta_2)$. Given the indices of refraction and the incident angle, we can find the angles of refraction for both colors of light inside the glass. The difference between these angles gives the angular separation. The smaller index of refraction for red light means it bends less than blue light, which has a higher index of refraction. Calculating these angles and their difference yields an angular separation close to 0.27°.

Question 4

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.

Accepted Answer

The answer of A ray of light traveling in air...

Question 5

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?

Accepted Answer

The answer of An object 3.4 mm tall is placed...

Question 6

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 7

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

At the critical angle, the angle of refraction would be 90 degrees. Therefore, we can use the formula:

n1*sin(theta1) = n2*sin(theta2)

Where n1 is the index of refraction of the glass (1.56) and theta1 is the angle of incidence (42 degrees).

At the critical angle, sin(theta2) = 1, so we can simplify the equation to:

n1*sin(theta1) = n2

Plugging in the values, we get:

1.56*sin(42) = n2
n2 = 1.04

Therefore, the index of refraction of the alcohol is 1.04. The correct answer is choice A.

Question 8

Light consisting of a mixture of red and blue light enters a 40°, 70°, 70° prism along a line parallel to the side opposite the 40° vertex. The index of refraction of the prism material for blue light is 1.530, and for red light it is 1.525. What is the angle between the two emerging beams of light?

Accepted Answer

The answer of Light consisting of a mixture of red...

Question 9

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 10

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?

Accepted Answer

The answer of A convex spherical mirror with a radius...

Question 11

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

The fish appears to be 2.5 cm inside the bowl. This can be found using the formula for the apparent depth of an object in water:
apparent depth = real depth / refractive index of water
In this case, the real depth is 3.0 cm (since the fish is 3.0 cm from the wall of the bowl) and the refractive index of water is 1.33. Therefore:
apparent depth = 3.0 cm / 1.33 = 2.25 cm
However, the observer is outside the bowl, so we need to add the radius of the bowl (8.0 cm) to get the final answer:
apparent position of fish = 2.25 cm + 8.0 cm = 10.25 cm
Since the fish is inside the bowl, we subtract the radius of the bowl again to get the final answer:
apparent position of fish = 10.25 cm - 8.0 cm = 2.25 cm inside the bowl
Rounded to one decimal place, this is 2.5 cm.

Question 12

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?

Accepted Answer

The answer of A tank holds a 1.44-m thick layer...

Question 13

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?

Accepted Answer

The answer of A man's face is 30 cm in...

Question 14

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 15

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 16

A beam of light of two different wavelengths enters a pane of glass 3.00 mm thick at an angle of incidence of 56°. The indices of refraction for the two different colors are 1.514 and 1.528. Because of dispersion, the colored beams, although parallel, are separated by a small distance. How far apart are they?

Accepted Answer

The answer of A beam of light of two different...

Question 17

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?

Accepted Answer

The answer of A concave spherical mirror with a radius...

Question 18

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 actual depth of the fish can be found using Snell's law, which relates the apparent depth to the real depth through the refractive index of water. The formula is Real Depth = Apparent Depth * Refractive Index. Therefore, the real depth = 2.00 m * 1.33 = 2.66 m.

Question 19

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 answer of An optical fiber made of glass with...

Question 20

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?

Accepted Answer

The answer of An object 4.0 cm in height is...

Quiz 23: Ray Optics