Chapter 23, Problem 39QAP

To determine

(a)

The angle of refraction ${\theta }_2$from the given diagram

Answer to Problem 39QAP

The angle of refraction ${\theta }_2=13.2°$

Explanation of Solution

Given:

Formula used:

Apply the refraction condition from Snell's Law,

  $n_1\sin {\theta }_1=n_2\sin {\theta }_2$

Here, all alphabets are in their usual meanings.

Calculation:

Apply the refraction condition from Snell's Law,

  $n_1\sin {\theta }_1=n_2\sin {\theta }_2$

  $\begin{array}{l}or,\sin {\theta }_2=\frac{n_1\sin {\theta }_1}{n_2}\\ or,{\theta }_2={\sin }^{-1}\left(\frac{n_1\sin {\theta }_1}{n_2}\right)\\ or,{\theta }_2={\sin }^{-1}\left(\frac{1.0\times \sin 20°}{1.50}\right)\\ or,{\theta }_2=13.2°\end{array}$

Hence, the angle of refraction ${\theta }_2=13.2°$

Conclusion:

Thus, the angle of refraction ${\theta }_2=13.2°$

To determine

(b)

The angle of refraction ${\theta }_2$from the given diagram

Answer to Problem 39QAP

The angle of refraction ${\theta }_2=22.1°$

Explanation of Solution

Given:

Formula used:

Apply the refraction condition from Snell's Law,

  $n_1\sin {\theta }_1=n_2\sin {\theta }_2$

Here, all alphabets are in their usual meanings.

Calculation:

Apply the refraction condition from Snell's Law,

  $n_1\sin {\theta }_1=n_2\sin {\theta }_2$

  $\begin{array}{l}or,\sin {\theta }_2=\frac{n_1\sin {\theta }_1}{n_2}\\ or,{\theta }_2={\sin }^{-1}\left(\frac{n_1\sin {\theta }_1}{n_2}\right)\\ or,{\theta }_2={\sin }^{-1}\left(\frac{1.0\times \sin 30°}{1.33}\right)\\ or,{\theta }_2=22.1°\end{array}$

Hence, the angle of refraction ${\theta }_2=22.1°$

Conclusion:

Thus, the angle of refraction ${\theta }_2=22.1°$

To determine

(c)

The angle of incident ${\theta }_1$from the given diagram

Answer to Problem 39QAP

The angle of incident is ${\theta }_1=30.9°$

Explanation of Solution

Given:

Formula used:

Apply the refraction condition from Snell's Law,

  $n_1\sin {\theta }_1=n_2\sin {\theta }_2$

Here, all alphabets are in their usual meanings.

Calculation:

Apply the refraction condition from Snell's Law,

  $n_1\sin {\theta }_1=n_2\sin {\theta }_2$

  $\begin{array}{l}or,\sin {\theta }_1=\frac{n_2\sin {\theta }_2}{n_1}\\ or,{\theta }_1={\sin }^{-1}\left(\frac{n_2\sin {\theta }_2}{n_1}\right)\\ or,{\theta }_1={\sin }^{-1}\left(\frac{1.5\times \sin 20°}{1.0}\right)\\ or,{\theta }_1=30.9°\end{array}$

Hence, the angle of incident is ${\theta }_1=30.9°$

Conclusion:

Thus, the angle of incident is

  ${\theta }_1=30.9°$

To determine

(d)

The angle of incident

  ${\theta }_1$from the given diagram

Answer to Problem 39QAP

The angle of incident is ${\theta }_1=22.8°$

Explanation of Solution

Given:

Formula used:

Apply the refraction condition from Snell's Law,

  $n_1\sin {\theta }_1=n_2\sin {\theta }_2$

Here, all alphabets are in their usual meanings.

Calculation:

Apply the refraction condition from Snell's Law,

  $n_1\sin {\theta }_1=n_2\sin {\theta }_2$

  $\begin{array}{l}or,\sin {\theta }_1=\frac{n_2\sin {\theta }_2}{n_1}\\ or,{\theta }_1={\sin }^{-1}\left(\frac{n_2\sin {\theta }_2}{n_1}\right)\\ or,{\theta }_1={\sin }^{-1}\left(\frac{1.5\times \sin 15°}{1.0}\right)\\ or,{\theta }_1=22.8°\end{array}$

Hence, the angle of incident is ${\theta }_1=22.8°$

Conclusion:

Thus, the angle of incident is

  ${\theta }_1=22.8°$