Chapter 23, Problem 37P

To determine

The term critical angle and to find the material in which light starts for its internally reflected case.

Answer to Problem 37P

Solution:

Since the refractive index of crown glass is higher than water so light must start from crown glass at its critical angle of $\angle c={61.0}^0$

Explanation of Solution

Given Info:

Critical Angle:

Angle of incident angle is said to be critical angle when its refracted ray makes right angle with the normal provided light inters from denser medium to rarer medium.

Refractive index of water $n_{air}=1.33$

Refractive index of crown glass $n_c=1.52$

Let the angle of incident (critical angle)be $\angle i=\angle c$

Angle of refraction be $\angle r={90}^0$

Formula Used:

Apply Snell’s Law

$\begin{array}{l}{n}_c\sin \angle i=n_{water}\sin \angle r\\ n_c\sin \angle c=n_{water}\sin {90}^0\end{array}$

Here, all alphabets are in their usual meanings.

Calculation:

Apply Snell’s Law

$\begin{array}{l}{n}_c\sin \angle i=n_{water}\sin \angle r\\ n_c\sin \angle c=n_{water}\sin {90}^0\\ or,\sin \angle c=\frac{1.33\sin {90}^0}{1.52}\\ or,\angle c={61.0}^0\end{array}$

Since the refractive index of crown glass is higher than water so light must start from crown glass at its critical angle of $\angle c={61.0}^0$