Optics Problems: Refraction and Lenses, Exams of Physics

Solutions to various optics problems related to refraction and lenses. The problems involve analyzing the behavior of light rays passing through different media and determining the indices of refraction, focal lengths, and image distances. Understanding these concepts is essential for studying optics and related fields.

Typology: Exams

Pre 2010

Uploaded on 02/10/2009

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CTRay-1 A man is looking in a mirror and he sees his faces just filling the mirror.
The man now moves back, away from the mirror, watching his reflection. As he moves
back, he sees the image of his face..
A: continues to just fill the mirror
B: becomes smaller than the frame of the mirror
C: become larger than the frame of the mirror.
Answer: From the man's point of view, his image always just fills the mirror. From his
point of view, both the mirror frame and the (virtual) image get smaller by the same
amount . If he doubles the distance L to the mirror, he is also doubling the distance 2L to
his image.
Man Virtual Image
pf3
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CTRay- 1 A man is looking in a mirror and he sees his faces just filling the mirror.

The man now moves back, away from the mirror, watching his reflection. As he moves

back, he sees the image of his face..

A: continues to just fill the mirror

B: becomes smaller than the frame of the mirror

C: become larger than the frame of the mirror.

Answer: From the man's point of view, his image always just fills the mirror. From his

point of view, both the mirror frame and the (virtual) image get smaller by the same

amount. If he doubles the distance L to the mirror, he is also doubling the distance 2L to

his image.

Man Virtual Image

CTRay- 2 A ray of light passes through 3 regions labeled I, II, and III, as shown. How do

the indices of refraction of regions I and III compare?

A: nI > nIII B: nI < nIII C: nI = nIII D: Impossible to tell.

Answer: nI < nIII. In refraction, the ray always gets bent toward the normal in the

medium with the larger n. The presence of medium II in between I and III has no effect

on the final ray's angle. If you shrink region II to a smaller slab, none of the angles

change. So you can shrink II down to zero thickness and ignore it.

CTRay- 3 A ray of light passes thru a sheet of glass which is thick at the bottom and thin

at the top. Which way is the ray traveling after it has passed through the glass?

A: bent toward the thin end B: undeviated C: bent toward the thick end

I II

III

o

o

o

o

air, n = 1

glass

n = 1.

A

B

C

A) Unchanged. B) greater, f > 20 cm. C) smaller, f < 20cm, but still

positive D) negative.

Answer: greater, f > 20cm. The rays of light are bent when they pass from the medium

(air or water) into the glass according to Snell's law. The amount of bending (the change

in the angle, ) depends on the change in the index of refraction, n. Bigger change in

n means bigger change in angle, , that is, more bending. If the lens is immersed in a

fluid with the same n as the glass, then there is no refraction: n = 0,  = 0. When the

lens is in air the change in n is n = 1.6 - 1 = 0.6. When the lens is in water, the change

in n is n = 1.6 - 1.33 = 0.27. Smaller n in the water means less bending, a longer focal

length f.

CTRay- 6 A bundle of parallel rays approaches the eye and some of the rays enter the

eye's pupil, as shown below. No other rays enter the eye. What does the eye see?

A) A single point of light, surrounded by blackness.

B) A uniformly illuminated wall of light, like a white wall.

C) Many scattered points of light, like stars in the night sky.

D) None of these

Answer: A single point of light, surrounded by blackness. A point source at infinity

makes a parallel bundle of rays. The lens of the eye focuses these rays onto a point on

the retina, and the brain perceives a point of light.

Eye

rays in water (less bending of ray, longer f)

rays

in air

CTRay- 7 Two point sources of light are imaged onto a screen by a converging lens as

shown. The images are labeled 1 and 2. A mask is used to cover up the left half of the

lens, as shown. What happens to the images on the screen when the mask is inserted over

the left half the lens?

A) Image 1 on left vanishes

B) Image 2 on right vanishes

C) Something else happens.

Answer: Something else happens. Both images dim somewhat but neither disappears.

Rays from each source cover the entire lens. When half the lens is covered, half the rays

from each source are blocked, but the other half get through, producing a dimmed image.

mask

lens

screen

(^1 )

screen

lens

mask

1 2

CTRay- 9 An object is placed is placed near a diverging lens, but the object is further

from the lens than the absolute value of the focal length of the lens. The image formed

is.. A) real B) virtual C) there is no image.

The magnitude of the image distance, compared to the object distance, is ...

A) smaller. B) greater.

Answers: The image is virtual and the image distance is smaller than the object distance.

The only way to understand this is to draw a ray diagram:

You get the same answers whether or not the object distance is further from the lens than

the focal point.

CTRay- 10

Is the image on a movie screen real or virtual? A) real B) virtual

Is the image seen with a virtual reality headset real or virtual? A) real B) virtual

optic axis object

image

Answers: Movie screen image is real. When the rays actually converge at a point in

space (where you can put a screen to see the image) then the image is real.

Virtual reality image is virtual. The image must be further than 25cm from the viewer's

eye for the viewer to be able to focus on the image.

CTRay- 11 An object is placed closer to a magnifying glass than the focal length.

image

This person needs eyeglasses with lenses that are

A) converging B) diverging C) either converging or diverging, depending on

much correction is needed.

Answer: diverging lens is needed. To be diverging, a lens must be

thinner in the middle. Eyeglasses to correct myopia are shaped like

this.

The job of the eyeglasses is to take the real image at do =  and

create a (virtual) image close enough so the myopic eye can focus

on it.

In my case, my myopia is so bad that I can focus on objects no

further than 11 cm from my eye. So for my glasses, when the object is at infinity ( do =

), the virtual image must be 11 cm from my eye, which means it must be about 9.5cm

from the glasses ( di = -9.5 cm = -0.095 m). The lens equation says:

o i

10.4 diopters

d d 0.095 m f

Diopters = 1/f, where f is in meters. (Optometrists always indicate lens power in

diopters.)

CTRay- 13 An object is near a mirror. The virtual image formed by the mirror is upright,

is smaller than the object, and is nearer the mirror than the object as shown. Is the mirror

A: complex B: concave  or C: convex ? (Don't vote A)

f too short

Hint: For a plane mirror, the (virtual) image is the same distance from the mirror as the

object. Imagine bending the mirror. Do the reflected rays bend so the image gets further

or closer to the mirror.

Answer: Convex. A convex mirrors bends the reflected rays more outward, which has

the effect of moving the image closer to the mirror

object

image

mirror

(B)

concave

(C)

convex

object

image