Light is incident normally on the short face of a 30°-60°-90° prism (Fig. P33.50). A drop of liquid is placed on the hypotenuse of the prism. If the index of refraction of the prism is 1.56, find the maximum index that the liquid may have for the light to be totally reflected.

We are given -30°-60°-90° as shown in the above figure. The refractive index of an optical material is expressed by n and represents the speed of light in the material. The prism is dropped by a drop of liquid and the ray is totally reflected; therefore, the incident angle is 90degree as the ray is incident normally. Snell’s is given by the equation

$n_a\sin {\theta }_a=n_b\sin {\theta }_b$

Where ${\theta }_a$represents the critical angle which is 60°. $n_a$is the refractive index of the liquid. Now, solve Snell’s law for and substitute the values for $n_a,{\theta }_a,{\theta }_b$

$$\begin{gathered} n_b\&=\frac{n_a\sin {\theta }_a}{\sin {\theta }_b} \\ =\left(1.56\right)\left(\frac{\sin 60°}{\sin 90°}\right) \\ =1.35 \end{gathered}$$

Hence, the maximum index is 1.35 that the liquid must have for the light to be totally reflected.