Question 1

A positive lens has a focal length f.The image is the same size as the object when
A)the object is at the focal point.
B)the image is on the opposite side of the lens from the object and is the same distance from the lens as the object.
C)the image is on the same side of the lens as the object and is the same distance from the lens as the object.
D)The image can never be the same size as the object.
E)None of these is correct.

Accepted Answer

By the lens formula, when the object distance is equal to the focal length, the image will be formed at infinity. Therefore, to have the same size image, the image distance should be equal to the object distance, which is not possible in option A. Option C is not possible as the image will always be inverted and on the opposite side of the lens. Option D is not correct as we have already found a valid option.

Question 2

A circular coin is 33.3 cm from a concave spherical mirror.(Assume the coin is on the optical axis.)The image of the coin is produced 10.0 cm from the mirror and on the same side as its real position.If the coin is moved to a position 23.3 cm from the mirror,calculate how far from the mirror the new image is:
A)13.7 cm
B)10.0 cm
C)15.1 cm
D)8.35 cm
E)11.5 cm

Accepted Answer

Using the mirror equation (1/f = 1/do + 1/di), we first find the focal length with the initial conditions (do = 33.3 cm, di = 10.0 cm). Then, using the new object distance (do = 23.3 cm), we solve for the new image distance (di), which is approximately 11.5 cm.

Question 3

A real object is 42 cm from a negative lens with f = -21 cm.The image is
A)real and 14 cm from the lens.
B)real and 42 cm from the lens.
C)virtual and 42 cm from the lens.
D)virtual and 21 cm from the lens.
E)virtual and 14 cm from the lens.

Accepted Answer

Using the lens formula 1/f = 1/v - 1/u, where f is the focal length, v is the image distance, and u is the object distance. Substituting f = -21 cm and u = -42 cm (since the object is on the opposite side of the light's direction, it's considered negative), we solve for v. The calculation shows the image is virtual and 14 cm from the lens on the same side as the object.

Question 4

When you stand in water up to your knees,your feet appear
A)closer than usual.
B)farther away than usual.
C)at the same location as usual.
D)To answer this you must know your height and the depth of the water.
E)To answer this you must know the index of refraction of water.

Accepted Answer

The answer of When you stand in water up to...

Question 5

Two parallel mirrors are a distance d apart.An object is placed 1.0 inch in front of the right-hand mirror.After 5 reflections,with the first reflection from the right-hand mirror,an image is located 17 inches behind the right-hand mirror.Calculate the distance between the mirrors.
A)5.3 in
B)3.2 in
C)6.0 in
D)4.0 in
E)3.6 in

Accepted Answer

The distance between the mirrors is 4.0 inches. Each reflection adds 2d to the total distance traveled by the image. After 5 reflections, the total distance is 1 + 5(2d) = 17 inches. Solving for d gives d = 4.0 inches.

Question 6

An object is located 3 cm in front of a converging lens of focal length 2 cm.A diverging lens of -6 cm focal length is placed 3 cm on the other side of the converging lens.What is the final image distance with respect to the diverging lens?
A)$\infty$
B)2 cm
C)-2 cm
D)6 cm
E)-6 cm

Accepted Answer

Using the thin lens formula, we can find the image distance with respect to the converging lens:
1/f_total = 1/f1 + 1/f2 - d/f1f2
1/f_total = 1/2 + (-1/6) - (3/12)
1/f_total = 1/3
f_total = 3 cm
Now using the formula for image distance with respect to the diverging lens: 
1/s2 + 1/f2 = 1/f_total
1/s2 + 1/-6 = 1/3
1/s2 = 1/6 - 1/3
1/s2 = -1/6
s2 = -6 cm

Therefore, the final image distance with respect to the diverging lens is -6 cm.

Question 7

An object 2 cm high is placed 20 cm from a thin convex lens that has a focal length of 10 cm.Which of the following statements describes the image formed?
A)It is 2 cm high.
B)It is 20 cm from the lens.
C)It is inverted.
D)It is real.
E)All of these are correct.

Accepted Answer

The answer of An object 2 cm high is placed...

Question 8

A positive lens of focal length 40 cm is placed 20 cm behind a negative lens of focal length -80 cm.An object is placed 240 cm in front of the negative lens.The final image distance relative to the positive lens is
A)-40 cm
B)56 cm
C)-56 cm
D)80 cm
E)-80 cm

Accepted Answer

The answer of A positive lens of focal length 40...

Question 9

A thin converging lens made of Crown glass (n = 1.517)is flat on one side and has a radius of curvature of 50 cm on the other side.An object is placed 45.0 cm to the left of this lens.Where is the image?
A)120 cm to the right of the lens
B)31 cm to the left of the lens
C)31 cm to the right of the lens
D)84 cm to the right of the lens
E)84 cm to the left of the lens

Accepted Answer

The answer of A thin converging lens made of Crown...

Question 10

A real object in front of a concave spherical mirror can produce an image that is
A)virtual,inverted,and magnified.
B)real,erect,and magnified.
C)diminished,erect,and virtual.
D)magnified,erect,and virtual.
E)diminished,real,and erect.

Accepted Answer

A concave spherical mirror can only produce virtual, erect and magnified images when the object is located between the mirror's focus and its center of curvature.

Question 11

The lens of a slide projector has a focal length of +12.0 cm.A slide 12.6 cm to the left of the lens forms the object for the lens.The screen should be placed
A)25 cm to the left of the lens.
B)25 cm to the right of the lens.
C)250 cm to the left of the lens.
D)250 cm to the right of the lens.
E)390 cm to the right of the lens.

Accepted Answer

The answer of The lens of a slide projector has...

Question 12

Which of the following statements is false?
A)The image produced by a diverging lens is always virtual,upright,and reduced in size.
B)The image produced by a converging lens can be virtual,upright,and magnified in size.
C)The image produced by a converging lens can never virtual,upright,and reduced in size.
D)To produce a real image,an object has to be place further away than the focal point of a converging lens.
E)The image produced by a converging lens cannot be real,inverted,and reduced in size.

Accepted Answer

The answer of Which of the following statements is false?
A)The...

Question 13

To project an image onto a screen using a lens,
A)the lens must be diverging and the object must be farther from the lens than the second focal point.
B)the lens must be converging and the object must be between the first focal point and the lens.
C)the lens must be diverging and the image must be farther from the lens than the second focal point.
D)the lens must be converging and the object must be farther from the lens than the first focal point.
E)the lens must be diverging and the object must be between the first focal point and the lens.

Accepted Answer

To project a real, inverted image onto a screen, a converging (convex) lens is used, and the object must be placed farther from the lens than the first focal point. This setup allows the lens to focus the light rays from the object onto a specific point on the screen, creating a clear image.

Question 14

A lens forms an erect image of a real object.The image is twice the size of the object and appears to be 40 cm from the lens.Determine the power of the lens in diopters (D).
A)There is not sufficient information.
B)-2.5 D
C)+2.5 D
D)+7.5 D
E)None of these is correct.

Accepted Answer

The lens forms an erect image of a real object, which means it is a diverging lens (concave lens). The image is virtual, erect, and magnified. The magnification (m) is given as 2 (since the image is twice the size of the object). The image distance (v) is given as -40 cm (negative because it's a virtual image on the same side of the lens as the object). Using the lens formula 1/f = 1/v + 1/u and the magnification formula m = -v/u, we can solve for the focal length (f). Since m = 2, v = -40 cm, solving these equations gives us f = -20 cm. The power (P) of a lens is given by P = 1/f (in meters), so P = 1/(-0.20) = -5 diopters. However, there seems to be a mistake in my calculation as none of the options match this result. Given the options, the closest approach to solving this with the correct method would lead to reconsidering the calculation or the interpretation of the given information. Since the provided answer does not match the options and considering the correct approach to calculate the power of a lens, there might have been an error in the calculation process. The correct answer should involve using the lens formula and magnification to find the focal length and then calculating the power, but without the specific correct calculations leading to one of the provided options, the response might have been based on an incorrect interpretation of the given data or a miscalculation.

Question 15

A real object is 42.1 cm from a negative lens that has a focal length of 25.6 cm.The object is 2.45 mm high.The height and orientation of the image are
A)3.8 mm and inverted.
B)0.93 mm and erect.
C)6.4 mm and erect.
D)0.38 mm and erect.
E)None of these is correct.

Accepted Answer

Using the lens formula 1/f = 1/do + 1/di, we can find the distance of the image from the lens as -16.56 cm (negative sign indicates an inverted image). Using the magnification formula M = -di/do = hi/ho, where hi and ho are the heights of the image and object respectively, we can find the height of the image as 0.93 mm (since M is negative, the image is also inverted). Therefore, the correct answer is B) 0.93 mm and erect, since the image is not only inverted but also smaller in size than the object, and appears on the same side of the lens as the object.

Question 16

A positive lens has a focal length f.The only way to get a magnification of -1 is to
A)place a real object at the focal point.
B)place a real object at a distance 2f from the lens.
C)place a virtual object at a distance 2f from the lens.
D)Magnifications can never be negative.
E)None of these is correct.

Accepted Answer

According to the lens formula, 1/f = 1/v - 1/u, where f is the focal length, v is the image distance, and u is the object distance. For a positive lens, u is positive for a real object, and v is positive for a real image. 
When the magnification m = -1, we have v/u = -1. 
Substituting this in the lens formula, we get:
1/f = 1/v + m/u = 1/u - 1/u = 0
Therefore, u = infinity, or the object must be at the far point of the lens. 
For a positive lens, the far point is at a distance 2f. Therefore, the only way to get a magnification of -1 is to place a real object at a distance 2f from the lens.

Question 17

A real object is 30 cm from a negative lens that has a focal length f = -20 cm.The image is
A)virtual and 30 cm from the lens.
B)real and 12 cm from the lens.
C)virtual and 12 cm from the lens.
D)real and 20 cm from the lens.
E)None of these is correct.

Accepted Answer

According to the lens formula,
1/f = 1/di - 1/do
where f = -20 cm (focal length of the lens),
do = 30 cm (object distance from the lens),
and di is the image distance from the lens.

Substituting the given values in the formula, we get
1/-20 = 1/di - 1/30
Solving this equation, we get
di = -12 cm

The negative value of the image distance indicates that the image is virtual, which means it is formed on the same side of the lens as the object.

Therefore, the answer is (C) virtual and 12 cm from the lens.

Question 18

A circular coin is 33.3 cm from a concave spherical mirror.(Assume the coin is on the optical axis.)The image of the coin is produced 23.3 cm towards the mirror from its real position.Calculate the focal length of the mirror.
A)15.4 cm
B)7.69 cm
C)13.7 cm
D)27.4 cm
E)10.0 cm

Accepted Answer

Using the mirror formula (1/f = 1/v + 1/u), where f is the focal length, v is the distance of the image from the mirror, and u is the distance of the object from the mirror, we have:

1/f = 1/v + 1/u
1/f = 1/23.3 + 1/33.3
1/f = (33.3 + 23.3)/(23.3 x 33.3)
1/f = 56.6/776.89
f = 13.7 cm

Therefore, the focal length of the mirror is 13.7 cm, which is option C.

Question 19

A cat is watching a goldfish in a spherical fishbowl of radius 15.0 cm.The goldfish is 5.00 cm from the side of the bowl nearest the cat.How far from the side of the bowl does the fish appear to be to the cat? (For water,n = 1.33.)
A)7.5 cm
B)4.5 cm
C)4.1 cm
D)3.5 cm
E)2.5 cm

Accepted Answer

When light travels from water to air, it bends away from the normal. In order to find out where the fish appears, we need to find the apparent depth of the fish. Using Snell's Law, we can find the angle of incidence:

sin(i)/sin(r) = n(water)/n(air)
sin(i) = n(air)/n(water) * sin(r)
sin(i) = (1.00/1.33) * sin(arcsin(5/15))
sin(i) = 0.376
i = 22.3 degrees

The cat's line of sight passes through the apparent position of the fish, so we can draw a right triangle where 22.3 degrees is opposite the shortest side (the distance from the fish to the bowl), and the radius of the bowl is the hypotenuse. Let x be the distance from the fish to the side of the bowl.

sin(22.3) = x/15
x = 15*sin(22.3)
x = 4.1 cm
 
Therefore, the fish appears to be 4.1 cm from the side of the bowl to the cat. The answer is (C).

Question 20

The clear water of a lake is 1.5 m deep at the end of a dock.You look directly down from the end of the dock and see a rock on the bottom of the lake.How far below the surface does the rock appear to be? (For water,n = 1.33.)
A)2.0 m
B)1.5 m
C)1.1 m
D)0.75 m
E)1.8 m

Accepted Answer

Using the formula:
apparent depth = real depth / refractive index of the medium
where real depth is 1.5 m and refractive index of water is 1.33.
So, 
apparent depth = 1.5 / 1.33 = 1.13 m
Thus, the rock appears to be 1.1 m below the surface. Therefore, the best choice is C.

Quiz 32: Optical Images

A positive lens has a focal length f.The image is the same size as the object when

A real object is 42 cm from a negative lens with f = -21 cm.The image is

When you stand in water up to your knees,your feet appear

Two parallel mirrors are a distance d apart.An object is placed 1.0 inch in front of the right-hand mirror.After 5 reflections,with the first reflection from the right-hand mirror,an image is located 17 inches behind the right-hand mirror.Calculate the distance between the mirrors.

An object is located 3 cm in front of a converging lens of focal length 2 cm.A diverging lens of -6 cm focal length is placed 3 cm on the other side of the converging lens.What is the final image distance with respect to the diverging lens?

An object 2 cm high is placed 20 cm from a thin convex lens that has a focal length of 10 cm.Which of the following statements describes the image formed?

A positive lens of focal length 40 cm is placed 20 cm behind a negative lens of focal length -80 cm.An object is placed 240 cm in front of the negative lens.The final image distance relative to the positive lens is

A thin converging lens made of Crown glass (n = 1.517) is flat on one side and has a radius of curvature of 50 cm on the other side.An object is placed 45.0 cm to the left of this lens.Where is the image?

A real object in front of a concave spherical mirror can produce an image that is

The lens of a slide projector has a focal length of +12.0 cm.A slide 12.6 cm to the left of the lens forms the object for the lens.The screen should be placed

Which of the following statements is false?

To project an image onto a screen using a lens,

A lens forms an erect image of a real object.The image is twice the size of the object and appears to be 40 cm from the lens.Determine the power of the lens in diopters (D) .

A real object is 42.1 cm from a negative lens that has a focal length of 25.6 cm.The object is 2.45 mm high.The height and orientation of the image are

A positive lens has a focal length f.The only way to get a magnification of -1 is to

A real object is 30 cm from a negative lens that has a focal length f = -20 cm.The image is

A circular coin is 33.3 cm from a concave spherical mirror.(Assume the coin is on the optical axis.) The image of the coin is produced 23.3 cm towards the mirror from its real position.Calculate the focal length of the mirror.

A cat is watching a goldfish in a spherical fishbowl of radius 15.0 cm.The goldfish is 5.00 cm from the side of the bowl nearest the cat.How far from the side of the bowl does the fish appear to be to the cat? (For water,n = 1.33.)

The clear water of a lake is 1.5 m deep at the end of a dock.You look directly down from the end of the dock and see a rock on the bottom of the lake.How far below the surface does the rock appear to be? (For water,n = 1.33.)