University of Rochester
Department of Physics and Astronomy
Physics123, Spring 2012
Recitation 5 - Solutions
Conceptual Questions:
•
When you look down into a swimming pool (or a lake) are you likely to
overestimate or underestimate its depth? Explain. How does the apparent depth vary with
the viewing angle?
Underestimate. The light rays leaving the bottom of the pool bend away from the normal
as they enter the air, so their source appears to be shallower than it actually is. The
greater the viewing angle, the more the bending of the light and therefore the less the
apparent depth.
eye air Apparent depth •
The level of water in a clear, colorless glass can easily be observed with the naked
eye. The level of liquid helium in a clear glass vessel is very difficult to see with the
naked eye. Explain.
The water level in a clear glass is observable because light is refracted as it passes from
air to water to air. The index of refraction of liquid helium has a gas-like value (1.026 for
helium I) and this makes its surface so hard to see that floats of styrofoam are often used
to show where the surface is.
•
Why do astronomers looking at distant galaxies talk about looking backward in
time?
Light travels through a vacuum at speed of 3×108 m/s. Thus, an image we see from a
distant star must have been generated some time ago. For example, the star Altair is 16
light-years awa; if we look at an image of Altair today, we know only what was
happening 16 years ago.
•
Does the mirror equation (Eq. 32-2) hold for a plane mirror? Explain.
1 Yes. For a plane mirror, d o = − d i , since the object and image are equidistant from the
mirror and the image is virtual, or behind the mirror. The focal length of a plane mirror is
infinite, so the result of the mirror equation, Eq. 32-2, is
1
do
+
1
di
= 0 , or d o = − d i , as
expected.
Problem R5.1
a) Demonstrate that a prism of apex angle α and refractive index n (see figure below)
deflects a ray incident at an angle θ by an angle θ d (known as deviation angle) equal to
(
)
θ d = θ − α + sin −1[
n 2 − sin 2 θ sin α − sin θ cos α ] .
[Hint: Use Snell’s law twice at the two refracting surfaces of the prism.]
b) Comment the case n = 1.5, α = 45°, and θ = 0.
c) Assuming that α is small (thin prism) and θ is also small (paraxial approximation),
write the small angle approximation of the formula above.
d) A parallel beam of light containing two wavelengths λ1 = 465 nm and λ2 = 652 nm
enters the prism at θ = 45.0°. Assume α = 60.0° and the prism made of silicate flint glass
(use table below). At what angle does each beam leave the prism?
e) In the small angle approximation considered in (c), write the angular dispersive power
of the prism, that is, dθ d d λ .
f) It can be demonstrated mathematically that for a given α and n , there is a minimum
deviation angle θ d,min that takes place when the passage of the rays through the prism is
symmetrical. That is, when the incident angle θ is equal to the angle at which the beam
leaves. Propose a physical argument that demonstrates the thesis above.
[Hint: If we reverse a light path, the ray will retract exactly the same path.]
θ
α
n
θd
2 Problem R5.1 - Solution
a) Defining
θ1 - the angle of incidence (θ in figure) from air to prism,
θ2 - the angle of refraction from air to prism,
θ3 - the angle of incidence from the prism to air,
θ4 - the angle of refraction from the prism to air,
θd – the deviation angle,
we have
θ2 + θ3 = α
θ1 + θ 4 = θ d + α .
and
Assuming the index of refraction in air equal to one and applying Snell’s law, we find
θ 4 = sin −1 [ n sin(α − θ 2 )] .
Upon expanding this expression, replacing the cosine and applying again Snell’s law we
have
θ 4 = sin −1 ⎡(sin α ) n 2 − sin 2θ1 − sin θ1 cos α ⎤ ,
⎣
⎦
from which we find the deviation angle θ d = θ1 − α + θ 4 ,
(
θ d = θ − α + sin −1[
)
n 2 − sin 2 θ sin α − sin θ cos α ] .
b) When n = 1.5, α = 45°, and θ = 0, the incident ray is totally reflected.
c) If the analysis is restricted to small angles:
⎧ θ1 ≅ nθ 2
⎪
⎨
⎪ nθ ≅ θ
3
4
⎩
⇒
θ d ≅ (n − 1)α .
d) The indexes of refraction for silicate flint glass are estimated from the graph as 1.642
for 465 nm and 1.619 for 652 nm. Each beam leaves the prism at θ 4 = θ d + α − θ1 . Using
the formula for the deviation angle we have
3 λ1 = 465 nm
→
θ 4 = 68.4 
λ1 = 652 nm
→
θ 4 = 65.2 .
e) The angular dispersive power in the small angle approximation (i.e. thin prism and
paraxial approximation) is
dθ d
dn
.
≅α
dλ
dλ
f) Suppose a ray undergoes a minimum deviation ( θ d = θ d,min ) and θ 4 ≠ θ . Then if we
reverse the ray, it will retrace the same path, so θ d must be unchanged, that is, it must
still give minimum deviation. But this imply that there are two different incident angles
for which the deviation is minimum, and this cannot be true.
4