Lesson 4:
Prisms and
Critical Angles

Optics HW #4 due NEXT TUESDAY!!!
◦ Note the change…
Lesson 4:
Prisms and Critical Angles
IV.C.1. Reflection and Refraction
 Students should understand the principles of
reflection and refraction, so they can:
b) Show on a diagram the directions of reflected and
refracted rays.
c) Use Snell’s Law to relate the directions of the incident
ray and the refracted ray, and the indices of refraction of
the media.
Students will be able to
1.
use Snell’s law to predict the ray path of light through
a prism.
2.
use Snell’s law to determine the critical angle for a
given boundary.

Whiteboard optics
Light enters a prism as shown, and passes through the
prism.
a) Complete the path of the light through the prism, and show the
angle it will make when it leaves the prism.
b) If the refractive index of the glass is 1.55, calculate the angle of
refraction when it leaves the prism.
c) How would the answer to b) change if the prism were immersed
in water?
air
30o
glass
60o
Light enters a prism made of air from glass.
a)
b)
Complete the path of the light through the prism, and show the
angle it will make when it leaves the prism.
If the refractive index of the glass is 1.55, calculate the angle of
refraction when it leaves the prism.
30o
glass
air
60o



The smallest angle of incidence for which
light cannot leave a medium is called the
critical angle of incidence.
If light passes into a medium with a
lesser refractive index than the original
medium, it bends away from the normal
and the angle of refraction is greater
than the angle of incidence.
If the angle of refraction is > 90o,
the light cannot leave the medium.
This drawing reminds us
that when light refracts
from a medium with a
larger n into one with a
smaller n, it bends away
from the normal.
n2
n1
n 1 > n2
This shows light hitting a
boundary at the critical
angle of incidence,
where the angle of
refraction is 90o.
No refraction occurs!
Instead of refraction,
total internal reflection
occurs when the angle
of incidence exceeds
the critical angle.
r = 90o
n2
c
n1
n 1 > n2
 n1sin(1)
= n2sin(90o)
 n1sin(c) = n2sin(90o)
 sin(c) = n2/ n1
 c = sin-1(n2/n1)
Binoculars use a
combination of
prisms that reflect
the incoming light.
As long as the
incident angles
exceed the critical
angle, the light will
be reflected.
Light enters the fiber
optic tube at an angle
above the critical angle
and is thus totally
reflected down the
‘light pipe’ to the other
end.
For commercial use,
two different glasses
are used, wrapped in a
protective cover. Which
must have the greater
index of refraction, the
core or the cladding?
A.
B.
What is the critical angle of incidence for a gemstone
with refractive index 2.45 if it is in air?
If you immerse the gemstone in water (refractive index
1.33), what does this do to the critical angle of
incidence?
The glass core of an optical fiber has an index of
refraction of 1.60. The index of refraction of the
cladding is 1.48.
What is the maximum angle a light ray can make
with the wall of the core if it is to remain inside the
fiber?