Question 1

The purpose of prescription glasses for a near-sighted person is to bring the apparent positions of distant objects

Accepted Answer

The purpose of prescription glasses for a near-sighted person is to bring the apparent positions of distant objects to the far point of the eye, which is the farthest point at which the eye can focus on an object without straining.

Question 2

An object is placed 25 cm in front of a lens of focal length 20 cm. 60 cm past the first lens is a second lens of focal length 25 cm. How far past the 25 cm lens does the final image form?

Accepted Answer

The answer of An object is placed 25 cm in...

Question 3

The object in a ray diagram is perpendicular to the principal axis but does not have one end lying on the principal axis. The minimum number of rays that must be drawn in this ray diagram in order to determine the position of the image is

Accepted Answer

The minimum number of rays that must be drawn in this ray diagram in order to determine the position of the image is 3. One ray should be drawn parallel to the principal axis, one ray should be drawn through the center of curvature, and one ray should be drawn through the focal point.

Question 4

A far-sighted student has a near point of 1.0 m. Calculate the focal length (in cm) of the glasses needed so the near point will be normal (25 cm).

Accepted Answer

We use the formula for lens-maker's equation:

1/f = (n-1) * (1/R1 - 1/R2)

Where f is the focal length, n is the refractive index of the lens material (we assume it to be 1.5), R1 is the radius of curvature of the first surface of the lens, and R2 is the radius of curvature of the second surface.

Since we want the near point to be 25 cm, the lens should create a virtual image of the object at 25 cm from the eyes. This means the lens behaves like a magnifying glass.

Using the formula for magnification, we have:

m = -d/f

Where d is the distance of the object (which is the near point) before the lens. The minus sign indicates that the image is virtual, which matches our requirement.

Plugging in the given values, we have:

d = 1.0 m (since the near point is 1.0m)

m = -25 cm / 1.0 m = -0.25

Substituting these values in the magnification formula, we get:

-0.25 = -1.5 / f

Solving for f, we get f = -6 cm.

However, this focal length is negative which means we need a diverging lens to get the desired virtual image. We want a converging lens, so we take the absolute value of f:

|f| = 6 cm

But the answer choices are in cm, so we convert 6 cm to 60 mm.

Thus, the answer is: Choice C, 33 mm.

Question 5

When you stand in front of a convex mirror, the image you see is

Accepted Answer

When you stand in front of a convex mirror, the image you see is virtual because the light rays do not actually converge at a point behind the mirror. The image is also upright because the top of the object is reflected as the top of the image and vice versa.

Question 6

If a convex lens were made out of very thin clear plastic filled with air, and were then placed underwater where n = 1.33 and where the lens would have an effective index of refraction n = 1, the lens would act in the same way

Accepted Answer

A lens' behavior depends on the difference in the index of refraction between the lens material and the surrounding medium. In this case, the lens (n = 1) has a lower index of refraction than the surrounding water (n = 1.33), which would cause it to diverge light rays similar to a concave lens in air. Convex lenses converge light rays, but because the lens' refractive index is lower than the water's, it acts oppositely to its design in air, thus behaving like a concave lens.

Question 7

A contact lens is made of plastic with an index of refraction of 1.50. The lens has an outer radius of curvature of +2.0 cm and an inner radius of curvature of +2.5 cm. What are the focal length and the power of the lens?

Accepted Answer

The answer of A contact lens is made of plastic...

Question 8

The image of an object beneath the surface of a medium of refractive index n > 1 is seen in air by a person looking down on the surface. This image, formed by light rays leaving the flat refractive surface, is

Accepted Answer

When light passes from a denser to a rarer medium, the light refracts away from the normal. In this case, light travels from a medium with refractive index n (denser) to air (rarer). Therefore, the image formed by the light rays leaving the surface of the medium is virtual and closer to the viewer than the object.

Question 9

A camera has a converging lens with a focal length of 56 mm. If the f-number is 2.8, what is the diameter of the lens (in mm)?

Accepted Answer

The formula for f-number is: f-number = focal length / diameter of lens
We are given the focal length (f) = 56 mm and the f-number = 2.8
Plugging in the values, we get:
2.8 = 56 / diameter of lens
Solving for diameter of lens:
diameter of lens = 56 / 2.8 = 20 mm
Therefore, the answer is D.

Question 10

The inhabitants of a planet in another galaxy have their eyes at the exact center of their 4.0-m long bodies. How long must a plane mirror be for such a creature to be able to see all of its body in the mirror?

Accepted Answer

To see the entire body in a mirror, the mirror only needs to be half the height of the body because the eyes are at the center. Thus, for a 4.0-m long body, a 2.0-m mirror is sufficient.

Question 11

A compound microscope is made with an objective lens (f₀ = 0.900 cm) and an eyepiece (f_e = 1.10 cm). The lenses are separated by a distance of 10.0 cm. What is the angular magnification? (Assume the near point is 25.0 cm.)

Accepted Answer

The angular magnification (M) of a compound microscope is given by the formula M = (L - f₀) / f₀ * (25 / fₑ), where L is the distance between the lenses. Substituting the given values: M = (10 - 0.9) / 0.9 * (25 / 1.1) = 9.1 / 0.9 * 22.727 ≈ -253.

Question 12

A girl is sitting on the edge of a pier with her legs dangling over the water. Her soles are 80.0 cm above the surface of the water. A boy in the water looks up at her feet and wants to touch them with a reed. (n_water = 1.333.) He will see her soles as being

Accepted Answer

Due to refraction, objects appear higher than they actually are when viewed from a denser medium like water. The apparent height can be calculated using the formula: apparent height = actual height / refractive index. So, 80 cm / 1.333 ≈ 60 cm. Adding this to the actual height gives approximately 107 cm.

Question 13

Of the possible statements that parallel rays are brought to a focus by a 1)
Concave mirror,
2)
Convex mirror,
3)
Converging lens,
4)
Diverging lens,
The correct answer is

Accepted Answer

Only the concave mirror and the converging lens can bring parallel rays of light to a focus. The convex mirror and diverging lens cause the parallel rays to diverge further. Therefore, the correct choice is C (1 and 3).

Question 14

An object is placed 25 cm in front of a lens of focal length 20 cm. 60 cm past the first lens is a second lens of focal length 25 cm. What is the resulting magnification of the object in this setup?

Accepted Answer

First, we can use the thin lens equation for the first lens:
1/f = 1/di + 1/do
where f is the focal length, di is the distance between the object and the lens, and do is the distance between the image and the lens. Plugging in the given values, we get:
1/20 = 1/25 + 1/do
Solving for do, we get:
do = 100 cm

Now we can consider the second lens. Let's use the thin lens equation again, this time using the image from the first lens as the object for the second lens:
1/f = 1/di + 1/do
where f is the focal length, di is the distance between the object and the lens, and do is the distance between the image and the lens. Plugging in the given values, we get:
1/25 = 1/60 + 1/do
Solving for do, we get:
do = 37.5 cm

The overall magnification of the object is the product of the magnification from each lens:
m = (-do/ di) * (-do/ di)
Plugging in the values we found, we get:
m = (-37.5/ 25) * (-100/ 25)
m = 6
Note that the negative signs cancel out because the image is on the opposite side of the lens as the object in both cases. Therefore, the resulting magnification is 6, which corresponds to answer choice A.

Question 15

A magnifying glass has a convex lens of focal length 15 cm. At what distance from a postage stamp should you hold this lens to get a magnification of +2.0?

Accepted Answer

The answer of A magnifying glass has a convex lens...

Question 16

A fish is 80 cm below the surface of a pond. What is the apparent depth (in cm) when viewed from a position almost directly above the fish? (For water, n = 1.33.)

Accepted Answer

The apparent depth of an object in water is given by the formula:

apparent depth = real depth / refractive index of water

In this case, the real depth is 80 cm and the refractive index of water is 1.33. So, the apparent depth is:

apparent depth = 80 cm / 1.33 = 60 cm

Therefore, the answer is Choice B, 60.

Question 17

A 10-cm focal length converging lens is used to magnify small newspaper print 2.0 mm high. Calculate the height of the image (in mm) for the maximum magnification for a normal eye.

Accepted Answer

For maximum magnification for a normal eye, the image should be at the near point of the eye, typically about 25 cm (or 250 mm) from the eye. The lens equation is 1/f = 1/v - 1/u, where f is the focal length, v is the image distance, and u is the object distance. For maximum magnification, v = 250 mm. The magnification (M) is given by M = v/u = h'/h, where h' is the image height and h is the object height. Rearranging the lens equation and solving for u gives u = fv/(v-f). Substituting f = 10 mm and v = 250 mm gives u ≈ 10.4 mm. The magnification is then M = 250/10.4 ≈ 24. Thus, the height of the image h' = Mh = 24 * 2 mm = 48 mm. However, since this option is not available, it seems there was a mistake in my calculation. The correct approach involves using the magnification formula directly related to the lens's ability to magnify objects for a normal eye, which typically involves considering the image formed at the near point for maximum magnification. Given the error in the initial calculation, let's correct the approach by considering the magnification formula M = -v/u, where the negative sign indicates that the image formed is inverted. For a lens to provide maximum magnification for a normal eye, the image should ideally be formed at the near point of the eye, which is generally considered to be 25 cm (or 250 mm) from the eye. The magnification is also given by the ratio of the image height to the object height (M = h'/h). Without the specific details on the calculation leading directly to the choices provided, the correct selection based on standard options would involve understanding that the magnification process depends on the object's distance from the lens and the image distance from the lens, which correlates with the eye's near point for maximum visibility.

Question 18

A compound microscope is made with an objective lens (  = 0.90 cm) and an eyepiece (  = 1.1 cm). The lenses are separated by a distance of 10 cm. If an object is 1.0 cm in front of the objective lens, where will the final image of the eyepiece be located?

Accepted Answer

To find the position of the final image, we need to use the formula for the magnification of a compound microscope:

M = (L / f1) × (1 + d / f2)

where 
L = distance between the lenses (10 cm)
f1 = focal length of objective lens (0.90 cm)
d = distance of object from objective lens (1.0 cm)
f2 = focal length of eyepiece (1.1 cm)

Plugging in the values and solving for the position of the final image, we get:

M = (10 / 0.90) × (1 + 1 / 1.1)
M = 13.95

Since the magnification is positive, the final image is upright. The position of the final image is given by:

di = (M × f1) + (d - f1)
di = (13.95 × 0.90) + (1.0 - 0.90)
di = 11.55 cm

Therefore, the final image of the eyepiece will be located 11.55 cm behind the eyepiece, which is equivalent to a distance of -11 cm relative to the objective lens. Hence, the answer is D (-11).

Question 19

A telescope is constructed with two lenses separated by a distance of 25 cm. The focal length of the objective is 20 cm. The focal length of the eyepiece is 5 cm. Calculate the magnitude of the angular magnification of the telescope.

Accepted Answer

The answer of A telescope is constructed with two lenses...

Question 20

If you stand closer to a concave mirror than a distance of one focal length, the image you see is

Accepted Answer

When standing closer than one focal length to a concave mirror, the image seen is virtual and upright. This is because the reflected light rays do not actually converge to form a real image, but instead appear to diverge from a virtual image point behind the mirror.

Quiz 36: Image Formation

The purpose of prescription glasses for a near-sighted person is to bring the apparent positions of distant objects

An object is placed 25 cm in front of a lens of focal length 20 cm. 60 cm past the first lens is a second lens of focal length 25 cm. How far past the 25 cm lens does the final image form?

The object in a ray diagram is perpendicular to the principal axis but does not have one end lying on the principal axis. The minimum number of rays that must be drawn in this ray diagram in order to determine the position of the image is

A far-sighted student has a near point of 1.0 m. Calculate the focal length (in cm) of the glasses needed so the near point will be normal (25 cm) .

When you stand in front of a convex mirror, the image you see is

If a convex lens were made out of very thin clear plastic filled with air, and were then placed underwater where n = 1.33 and where the lens would have an effective index of refraction n = 1, the lens would act in the same way

A contact lens is made of plastic with an index of refraction of 1.50. The lens has an outer radius of curvature of +2.0 cm and an inner radius of curvature of +2.5 cm. What are the focal length and the power of the lens?

The image of an object beneath the surface of a medium of refractive index n > 1 is seen in air by a person looking down on the surface. This image, formed by light rays leaving the flat refractive surface, is

A camera has a converging lens with a focal length of 56 mm. If the f-number is 2.8, what is the diameter of the lens (in mm) ?

The inhabitants of a planet in another galaxy have their eyes at the exact center of their 4.0-m long bodies. How long must a plane mirror be for such a creature to be able to see all of its body in the mirror?

A compound microscope is made with an objective lens (f₀ = 0.900 cm) and an eyepiece (f_e = 1.10 cm) . The lenses are separated by a distance of 10.0 cm. What is the angular magnification? (Assume the near point is 25.0 cm.)

A girl is sitting on the edge of a pier with her legs dangling over the water. Her soles are 80.0 cm above the surface of the water. A boy in the water looks up at her feet and wants to touch them with a reed. (n_water = 1.333.) He will see her soles as being

Of the possible statements that parallel rays are brought to a focus by a 1) Concave mirror, 2) Convex mirror, 3) Converging lens, 4) Diverging lens, The correct answer is

An object is placed 25 cm in front of a lens of focal length 20 cm. 60 cm past the first lens is a second lens of focal length 25 cm. What is the resulting magnification of the object in this setup?

A magnifying glass has a convex lens of focal length 15 cm. At what distance from a postage stamp should you hold this lens to get a magnification of +2.0?

A fish is 80 cm below the surface of a pond. What is the apparent depth (in cm) when viewed from a position almost directly above the fish? (For water, n = 1.33.)

A 10-cm focal length converging lens is used to magnify small newspaper print 2.0 mm high. Calculate the height of the image (in mm) for the maximum magnification for a normal eye.

A compound microscope is made with an objective lens ( = 0.90 cm) and an eyepiece ( = 1.1 cm) . The lenses are separated by a distance of 10 cm. If an object is 1.0 cm in front of the objective lens, where will the final image of the eyepiece be located?

A telescope is constructed with two lenses separated by a distance of 25 cm. The focal length of the objective is 20 cm. The focal length of the eyepiece is 5 cm. Calculate the magnitude of the angular magnification of the telescope.

If you stand closer to a concave mirror than a distance of one focal length, the image you see is

Quiz 36: Image Formation

The purpose of prescription glasses for a near-sighted person is to bring the apparent positions of distant objects

An object is placed 25 cm in front of a lens of focal length 20 cm. 60 cm past the first lens is a second lens of focal length 25 cm. How far past the 25 cm lens does the final image form?

The object in a ray diagram is perpendicular to the principal axis but does not have one end lying on the principal axis. The minimum number of rays that must be drawn in this ray diagram in order to determine the position of the image is

A far-sighted student has a near point of 1.0 m. Calculate the focal length (in cm) of the glasses needed so the near point will be normal (25 cm) .

When you stand in front of a convex mirror, the image you see is

If a convex lens were made out of very thin clear plastic filled with air, and were then placed underwater where n = 1.33 and where the lens would have an effective index of refraction n = 1, the lens would act in the same way

A contact lens is made of plastic with an index of refraction of 1.50. The lens has an outer radius of curvature of +2.0 cm and an inner radius of curvature of +2.5 cm. What are the focal length and the power of the lens?

The image of an object beneath the surface of a medium of refractive index n > 1 is seen in air by a person looking down on the surface. This image, formed by light rays leaving the flat refractive surface, is

A camera has a converging lens with a focal length of 56 mm. If the f-number is 2.8, what is the diameter of the lens (in mm) ?

The inhabitants of a planet in another galaxy have their eyes at the exact center of their 4.0-m long bodies. How long must a plane mirror be for such a creature to be able to see all of its body in the mirror?

A compound microscope is made with an objective lens (f0 = 0.900 cm) and an eyepiece (fe = 1.10 cm) . The lenses are separated by a distance of 10.0 cm. What is the angular magnification? (Assume the near point is 25.0 cm.)

A girl is sitting on the edge of a pier with her legs dangling over the water. Her soles are 80.0 cm above the surface of the water. A boy in the water looks up at her feet and wants to touch them with a reed. (nwater = 1.333.) He will see her soles as being

Of the possible statements that parallel rays are brought to a focus by a 1) Concave mirror, 2) Convex mirror, 3) Converging lens, 4) Diverging lens, The correct answer is

An object is placed 25 cm in front of a lens of focal length 20 cm. 60 cm past the first lens is a second lens of focal length 25 cm. What is the resulting magnification of the object in this setup?

A magnifying glass has a convex lens of focal length 15 cm. At what distance from a postage stamp should you hold this lens to get a magnification of +2.0?

A fish is 80 cm below the surface of a pond. What is the apparent depth (in cm) when viewed from a position almost directly above the fish? (For water, n = 1.33.)

A 10-cm focal length converging lens is used to magnify small newspaper print 2.0 mm high. Calculate the height of the image (in mm) for the maximum magnification for a normal eye.

A compound microscope is made with an objective lens ( = 0.90 cm) and an eyepiece ( = 1.1 cm) . The lenses are separated by a distance of 10 cm. If an object is 1.0 cm in front of the objective lens, where will the final image of the eyepiece be located?

A telescope is constructed with two lenses separated by a distance of 25 cm. The focal length of the objective is 20 cm. The focal length of the eyepiece is 5 cm. Calculate the magnitude of the angular magnification of the telescope.

If you stand closer to a concave mirror than a distance of one focal length, the image you see is

A compound microscope is made with an objective lens (f₀ = 0.900 cm) and an eyepiece (f_e = 1.10 cm) . The lenses are separated by a distance of 10.0 cm. What is the angular magnification? (Assume the near point is 25.0 cm.)

A girl is sitting on the edge of a pier with her legs dangling over the water. Her soles are 80.0 cm above the surface of the water. A boy in the water looks up at her feet and wants to touch them with a reed. (n_water = 1.333.) He will see her soles as being