CT0pt-1 A ray of light is bent when it passes from water to air.
air
water
This is an example of:
A) Huygen's Principle
C) The superposition principle
B) Snell's Law
D) The lens equation
Answer: Snell's Law
CT0pt -2. A man is looking in a mirror and he sees his face 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.
Hint:
Man
Virtual
Image
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
CT0pt -3 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?
II
I
III
25°
25°
40°
A) nI > nIII
35°
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.
I
II
III
35°
40°
CT0pt -4. 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 (goes straight thru)
C) bent toward the thick end
A)
B)
air, n=1
glass, n=1.5
C)
Answer: ASSUMING that the glass prism is surrounded by air, the ray is bent away from the thin end,
toward the thick end. On entering the glass, the ray is bent toward the normal (since the ray is always
bent toward the normal in the higher-n material). On exiting, it is bent away from the normal. Both
refractions, bend the ray away from the thinner end.
CT0pt -5. A light ray passes from air to glass (smaller n to larger n) as shown. Which way does the ray
travel in the glass?
A)
air, n=1
B)
C)
glass, n=1.5
D)
Answer: path C, the ray is entering a larger-n material, so the angle with the normal direction must be
smaller.
CT0pt -6. A light ray inside glass is totally internally reflected from an air-glass interface as shown.
air
glass
The air surrounding the glass is replaced with water. With the same light ray in the glass, the total
internal reflection will now...
A) definitely not occur.
B) definitely occur.
C) not enough information to know
Answer: not enough information to know. Whether total internal reflection will occur depends on the
angle of incidence and on the relative indices of refraction of the glass and water. The angle of
incidence must be greater than the critical angle which is given by:
n g sin c  n w sin 900  n w
n
c  sin 1  w
 ng




CT0pt -7. A converging lens has a focal length f = 20cm when it is in air. The lens is made of glass
with index of refraction nglass = 1.6. When the lens is placed in water (nwater = 1.33), the focal length of
the lens is..
..
A) Unchanged.
B) greater, f > 20 cm.
C) smaller, f < 20cm, but still positive
D) negative.
Answer: 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.
rays in water (less bending of ray, longer f)
rays
in air
CT0pt -8. 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?
Eye
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.
CT0pt -9 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?
1
2
screen
A) Image 1 vanishes
lens
B) Image 2 vanishes
mask
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.
1
2
screen
lens
mask
CT0pt -10 A converging lens focuses the light from a nearby point source onto an image, as shown.
The "focal point" of a lens is the point on the optic axis, one focal length f from the lens. Where is the
focal point of this lens?
A) between the lens and the image
B) at the image
C) further from the lens than the image.
A) B)
object
C)
image
lens
Answer: the focal point is between the image and the lens. The focal point is where parallel rays (along
the optic axis) would focus. A ray hitting the edge of the lens will be bent by same angle  , regardless
of the incoming angle. So when the object moves (from infinity) toward the lens, the image moves
away from the lens, but the focal point remains stationary by definition.
α
lens
α, same angle
CT0pt -11. 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:
object
image
optic axis
You get the same answers whether or not the object distance is further from the lens than the focal point.
CT0pt -11 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
Answers:
Question 1: 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. The image is Real.
Question 2: The image must be further than 25cm from the viewer's eye for the viewer to be able to
focus on the image. The image is Virtual.
image
Q0pt -12. An object is placed closer to a magnifying glass than the
focal length.
image
(virtual!)
optic axis
object
|di|
|do|
What are the signs of the focal length f, the object distance do,
and the image distance di?
A) f > 0, do > 0, di < 0.
B) f < 0, do > 0, di < 0.
C) f > 0, do < 0, di > 0.
D) f > 0, do > 0, di > 0.
If do = 5cm, |di| = 15 cm, and the object height ho is 1cm, what is the image height hi ?
A) 2cm B) 3cm C) 4cm
D) None of these.
Answers: Question 1: f > 0, do > 0, di < 0. f is positive for converging lenses (f is negative for diverging
lenses). Object distance is positive, do >0, for real objects on the left of lens (always the case, in this
course). Image distance is negative , di < 0, for virtual objects on the left of the lens.
Question 2: Image height is 3cm,
hi
d 15
 i  3
h o do 5
CT0pt -13. A person who is "near-sighted", or myopic, cannot focus on faraway objects (objects at
infinity) because the curvature of the person's eye-lens is too great. This causes parallel rays (from a
distant point source) to bend too much and focus in front of the retina. The person sees a fuzzy patch of
light rather than a sharp point.
f (too short)
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, see diagram below. 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 11cm 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:
1
1
1

 ,
d o di
f
1
1
1

  10.5 diopters
 0.095m f
Diopters = 1/f, where f is in meters. (Optometrists always indicate lens power in diopters.)
CT0pt -14 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 B: concave  or C:
convex  ?
mirror
object
image
(A)
concave
(B)
convex
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.
image
object
Answer: the mirror must be convex (B)
CT0pt -15 What's wrong with that scene in 2001?
A) Astronaut Dave Bowman can't read the screens on his console.
B) He's not weightless.
C) There's no thrust plume coming from his pod.
D) The background stars are moving.
Answer: Real images of the console displays are being projected onto his face, so Dave can't read the
screens on his console. This is an inaccuracy which was consciously inserted by the director for
cinematic effect; as far I can tell, this is the only physics error in the whole movie which was made
intentionally.
Ch0pt -16. An astronomical refracting telescope has 3/4 of its objective lens covered with a mask.
The observer reports that, compared to the image with no mask, the image is
A) unchanged.
B) 3/4 covered.
C) the same image, but 1/4 as bright.
D) the same image but 1/16 as bright.
Answer: the same image, but 1/4 as bright. Covering part of the lens still produces an image (see
concept test 9 above)
Shorter focal length magnifying glass give bigger virtual image.
smaller angle,
smaller image
bigger angle,
bigger image
Longer focal length lens produces larger real image.
far away
arrow
small image
small f
far away
arrow
big image
big f