A 45° - 45° - 90° prism is immersed in alcohol (n = 1.45). What is the minimum refractive index the prism must have if a ray incident normally on one of the short faces is to be totally reflected at the long face of a prism?

Mumbai University > Electronics Engineering > Sem7 > Optical Fiber Communication

Marks: 5M

Year: May 2012

Given:

$θ_c$= 45° $n_1$ = 1.45

To find: refractive index $n_2$

Solution:

$$θ_c= sin^{-1} \frac{n_2}{n_1} \\ \; \\ sin^{-1} \frac{n_2}{n_1} = 45° \\ \; \\ \frac{n_2}{n_1} = sin 45 = 0.707 \\ \; \\ n_2 = 1.45 * 0.707 \\ \; \\ = 1.025$$

$n_2$= Minimum refractive index= 1.025

$$\boxed{Minimum \ \ refractive\ \ index(n_2)= 1.025}$$