Concept Question
 = 47
47
43
47
47
4343
•This works for
any angle
•In 3D, you
need three
mirrors
Mirror
Mirror
A light ray starts from a wall at an angle
of 47 compared to the wall. It then
strikes two mirrors at right angles
compared to each other. At what angle
 does it hit the wall again?
A) 43 B) 45 C) 47 D) 49 E) 51
Concept Question
n1 sin 1  n2 sin 2
n5 = 1.5
6
n6 = 1
5
5
n3 = 2.4
n4 = 1.33
4 4
n1 = 1 n2 = 1.5
3
3
2
2
34
A light ray in air enters a region at an angle of 34.
After going through a layer of glass, diamond, water,
and glass, and back to air, what angle will it be at?
A) 34
B) > 34 C) <34
D) This is too hard
n1 sin 34  n2 sin 2  n3 sin 3  n4 sin 4  n5 sin 5  n6 sin 6
1 sin 34  1 sin 6
6  34
Concept Question
A trick question:
A light ray in diamond enters an air gap
at an angle of 30, then returns to
diamond. What angle will it be going
at when it leaves out the bottom?
A) 30
B) < 30
C) >30
D) None of the above
n1 sin 60  n2 sin 2
sin 2  2.4  0.5  1.2
•This is impossible!
•Light never makes it into region 2!
•It is totally reflected inside region 1
30
n1 = 2.4
2
n3 = 2.4
2
n2 = 1
3
Concept Question
•The image distance q is how far
the image is in front of the mirror
•Real image if q > 0, virtual
image if q < 0
•The magnification M is how large
the image is compared to the object
•Upright if positive, inverted if
negative
h
h’
Object
p
q Image
Mirror
h
M
h
If you place an object in front of a flat mirror,
its image will be
A) Real and upright B) Virtual and upright
C) Real and inverted D) Virtual and inverted
Concept Question
f  12 R
1 1 1
 
p q f
Light from the Andromeda Galaxy bounces off of a concave mirror with
radius R = 1.00 m. Where does the image form?
A) At infinity
B) At the mirror
C) 50 cm left of mirror
D) 50 cm right of mirror
•Concave, R > 0
f  12 R  50 cm
p  2 Mly  
1
1 1 1
0
  
q f p 50 cm
q  50 cm
Concept Question
A fish is swimming 24 cm underwater (n = 4/3). You are
looking at the fish from the air (n = 1). You see the fish
A) 24 cm below the water
B) 24 cm above the water
C) 32 cm below the water
D) 32 cm above the water
E) 18 cm below the water
1  24 cm 
q
43
 18 cm
24 cm
18 cm
•R is infinity, so formula above is valid
•Light comes from the fish, so the water-side is the front
p  24 cm
•Object is in front
•Light starts in water
n1  4 3
•For refraction, q tells you
distance behind the boundary n2  1
q
n2
p
n1
Using the Lens Maker’s Equation
•If you are working in air,
1  n2  1 1 
   1  
n1 = 1, and we normally call
f  n1  R1 R2 
n2 = n.
•By the book’s conventions, R1, R2 are positive if they
are convex on the front
•You can do concave on the front as well, if you use
negative R
•Or flat if you set R = 
If the lenses at
right are made of
A B
glass and are used
in air, which one
definitely has f < 0?
D
C
•If f > 0, called a
converging lens
•Thicker in middle
•If f < 0, called a
diverging lens
•Thicker at edge
•If you turn a lens
around, its focal length
stays the same
Light entering on the left:
•We want R1 < 0: first
surface concave on left
•We want R2 > 0: second
surface convex on left
Concept Question
What can you do if you don’t have perfect eyesight?
•To make the eye work, just put a lens that turns the object (p) you want
to see into an image at a distance (q) where you can see it
A farsighted person can’t see objects closer than 1.00 m away. What
focal length lens would adjust his eyesight so he can read 0.50 m away?
A) +1 m
B) -1 m
C) +3 m
D) -3 m
•The object will be 0.50 m in front of the lens
•p = +0.50 m
•The image will be 1.00 m in front of the lens
•q = -1.00 m
Opticians give the inverse
focal length, f -1, which is
given in diopters (= m-1)
1 1 1
 
p q f
1
1
1


0.5 m 1 m f
2 m1  1 m1 
1
f
f 1 m
Concept Question
A beam of light is to be put through a
small slit as shown at right. After it goes
through, which way will it spread out?
A) Horizontally
B) Vertically
C) An equal amount of both
•In vertical direction, waves add up to make a front going same direction
•In horizontal direction, waves spread out because slit is narrow
•Controlling the horizontal position causes its horizontal direction to
become uncertain
•Closely related to Heisenberg’s Uncertainty Principle!
Concept Question
If you used a little wider slit, the pattern would
A) Get wider and dimmer
B) Get wider and brighter
C) Get narrower and dimmer
D) Get narrower and brighter
 sin  a sin    
I  I max 


a
sin




2
Concept Question
What is the approximate ratio d/a of the
slit separation d over the slit size a?
A) 1 B) 2 C) 4 D) 8 E) 16 F) 32
Concept Question

2 dn
0
Light of wavelength 0.5 m takes two paths, both of length 1 m,
one through air, the other through glass (n = 1.5). What is the
difference in phase between the two waves in the end?
A) 0 B)  C) 2 D) 3 E) None of the above
1 m
2 11.5
1 
 6
0.5
2 11.0 
2 
 4
0.5
  2
Concept Question
Suppose we are in a glass medium, and we have a wave that goes
from glass to air to glass. If the layer of air is much smaller than
one wavelength, then the two reflected waves will add
A) Constructively
B) Destructively
C) Insufficient Info
•First transition: high to low
•no phase shift
•Second transition: low to high
• phase shift
•Compared to each other, the two waves are  out of phase with each other
•They will have a tendency to cancel
•Very little effect from layer if much thinner than a wavelength
Concept Question
If a wave is moving in the z-direction, in which directions can the
electric field point?
A) The x-direction (only)
B) The y-direction (only)
C) The z-direction (only)
D) The x- or y-direction, but not the z-direction
E) The x-, y-, or z-direction
Concept Question
A red laser and a blue laser each are emitting 12 mW of power.
Which one has a faster rate of photons coming out?
A) The red one
B) The blue one
C) It is a tie
D) Insufficient Information
Concept Question
h
   
1  cos  
mc
What is the maximum change that can occur to
the wavelength of light when it scatters from a
stationary electron?
A) Increase by h/mc B) Increase by 2h/mc
C) Decrease by h/mc D) Decrease by 2h/mc
•When it scatters forward,  = 0
   
h
h
1

cos




1  1
mc
mc
•When it scatters backwards,  = 
h

  
1  cos    h 1  1
mc
mc
  
2h
   
mc