Question

A triangular prism of glass is shown in the figure. A ray incident normally to one face is totally reflected. If $ \theta = {45^0} $ , the refractive index of glass is:

(A) $ < 1.41 $
(B) $ = 1.41 $
(C) $ > 1.41 $
(D) $ 1.732 $

Answer 1

Hint :If a ray is incident obliquely that is perpendicular to the glass surface then it moves into the second medium undeviated. Here ray is totally internally reflected; it is a special case in which light travels from a denser medium to a rarer medium.

Complete Step By Step Answer:
Initially the light is perpendicular to the surface, so it moves in a straight line. Then it is incident at an angle $ \theta = {45^0} $
We can use Snell’s law here;
Since light is internally reflected the angle $ \theta $ is greater than the critical angle.

The dashed line is perpendicular on side ae. In triangle abc, one angle is $ {45^0} $ and one angle is $ {90^0} $ since it is a right-angled triangle. So, by angle sum property the value of the third angle is $ {45^0} $ . Also, the incident ray makes an angle $ \theta $ , so $ \theta + {45^0} = {90^0} $
So $ \theta = {45^0} $
From Snell’s law, $ \mu = \dfrac{{\sin i}}{{\sin r}} $
Now for total internal reflection $ \theta > {\theta _c} $
 $
  \sin \theta > \sin {\theta _c} \\
   \Rightarrow \sin \theta > \dfrac{1}{\mu } \\
  \mu > \dfrac{1}{{\sin \theta }} \\
  \mu > 1.41 \\
  $
Hence, the correct answer is (C).

Additional Information:
For total internal reflection to take place two conditions must be satisfied. These are;
1-The light ray moves from a denser medium to less dense medium.
2-The angle of incidence must be greater than the critical angle

Note :
while using Snell’s law the refractive index is given by the ratio of refractive of the refractive index of the second medium to the refractive index of first medium. Be careful while the angles of reflection, refraction and of incidence are always made with the normal.