Webwork Problem #1 —- Assignment 11
The problem is to find θ such that total internal reflection is achieved at the
second and third air-prism interfaces. It is misleading, and I fell into this error
in class, of simply regarding the second interface and calculating the incident
angle prescribed by the total internal reflection law n sin θ = 1. The reason is
because I was sloppy and did not recall that this is the critical angle — any
angle greater produces the same effect. The second interface illustrates this
point.
Therefore, if the α is the incident angle at the second interface, the angle of
incidence at the third interface is 90◦ −α. Now, most of the webworks (if not all)
have an index of refraction resulting in an angle less than 45◦ . Therefore, of the
two interfaces, only the third can assume the critical angle, since α is necessarily
larger than this angle. Now, using some geometry we notice that the refracted
angle and α add to equal 45◦ (I leave it to you to see this). Therefore, if we
use Snell’s law and the law of total internal reflection and have α assume the
critical value θc , we have an equation for θ(n),
sin−1
θc + θr = 45◦ ,
µ
¶
1
sin θ
+ sin−1
= 45◦ .
n
n
Now it is a matter of algebra and trigonometry,
µ
¶
sin θ
−1 1
◦
= sin 45 − sin
,
n
n
Ãr
!
1
1
1
1− 2 −
=√
,
n
n
2
Ã√
!
n2 − 1 − 1
−1
√
=⇒ θ = sin
.
2
1