RayII-1. 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: 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.
rays in water (less bending of ray, longer f)
rays
in air
RayII-2 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
screen
2
A: Image 1 vanishes
B: Image 2 vanishes
lens
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
screen
2
lens
mask
RayII-3. 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.
RayII-4 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.
image
C
A
B
object

same angle
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.
RayII-5.
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.
object
image
optic axis
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.
RayII-6. An object is placed closer to a magnifying glass than the focal length.
image
(virtual!)
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:
f positive for converging lenses. Object distance positive, l >0, for real objects (always
the case, in this course). Image distance negative , l' < 0, since image is virtual and on
left of lens.
hi
d
5
 i 
3
ho
d o 15
RayII-7. 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.
Image height is 3cm.
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. To be diverging, a lens must be
thinner in the middle. Eyeglasses to correct myopia are shaped like:
The job of the eyeglasses is to take the real image at l =  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 (l = ), the virtual image must be 11 cm from my eye, which means it must be
about 9.5cm from the glasses (l' = -9.5 cm = -0.095 m). The lens equation says:
1 1 1
  ,
L L' 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.)
9.5cm
11.0 cm
RayII-9. 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.
E: None of these.
Answer: The image is (1/4) as bright, but is otherwise unchanged. (Actually, there will
also be some loss of resolution, but this will not be very apparent to the viewer, unless the
magnification is very high.)
RayII-10
What's wrong with that scene in 2001 where Astronaut Dave Bowman is piloting his
pod?
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.
E: Something else is wrong.
Answer: He can’t read the screens. The movie shows images of his display consoles
being projected onto his face, meaning there must be a converging lens between his
console and face. This was done for cinematic effect. But if there was a lens between
Dave and his console, he would be able to read the screen.