Chapter 18, Problem 51A

To determine

The critical angle for a special glass if the glass is immersed in water.

Answer to Problem 51A

  $61.05°$

Explanation of Solution

Given:

The critical angle for a special glass in air is $41.0°$

Formula used:

Critical angle is given as,

  ${\theta }_c={\sin }^{-1}\left(\frac{n_2}{n_1}\right)$

  $\begin{array}{l}{n}_1=\text{ index of refraction of incident medium}\\ n_2=\text{ index of refraction of refracting medium}\end{array}$

Calculation:

It is known that,

Index of refraction of air, $n{a}_{}=1.0003$

Index of refraction of water, $n{w}_{}=1.33$

Now, using the critical angle for a special glass in air, the refractive index of the glass will be,

  $\begin{array}{c}{\theta }_c={\sin }^{-1}\left(\frac{n_a}{n_g}\right)\\ 41°={\sin }^{-1}\left(\frac{1.0003}{n_g}\right)\\ n_g=1.52\end{array}$

Then, using the given and calculated values, the critical angle when glass is immersed in water will be,

  $\begin{array}{c}{\theta }_c={\sin }^{-1}\left(\frac{n_w}{n_g}\right)\\ ={\sin }^{-1}\left(\frac{1.33}{1.52}\right)\\ =61.05°\end{array}$

Conclusion:

Hence, the critical angle when glass is immersed in water is $61.05°$ .