Chapter 31, Problem 38P

(a)

To determine

The angle of incidence.

Answer to Problem 38P

The angle of incidence is ${\sin }^{-1}\left(n\sqrt{1\sqrt{1-\frac{1}{n^2}}}\right)$ .

Explanation of Solution

Formula used:

The expression for $b$ can be given as,

  $b=2d\tan {\theta }_r\cos {\theta }_i$ .

Here, $b$ is the separation between rays, $d$ is the thickness of plate ${\theta }_i$ is the angle of incidence and ${\theta }_r$ is the angle of refraction.

The expression for Snell’s law is given as,

  $\sin {\theta }_i=n\sin {\theta }_r$ …… (1)

Calculation:

The differentiation of $b$ can be calculated as,

  $\begin{array}{l}b=2d\tan {\theta }_r\cos {\theta }_i\\ \frac{db}{d{\theta }_i}=2d\left(-\tan {\theta }_r\sin {\theta }_i+{\sec }^2{\theta }_r{\cos }^2{\theta }_i\frac{d{\theta }_r}{d{\theta }_i}\right)\end{array}$ …… (2)

Differentiate equation (1).

  $\begin{array}{c}\sin {\theta }_i=n\sin {\theta }_r\\ \frac{d{\theta }_r}{d{\theta }_i}=\frac{\cos {\theta }_i}{n\cos {\theta }_r}\end{array}$ …… (3)

From equation (1), (2) and (3)

  $\frac{db}{d{\theta }_i}=\frac{2d}{n^3{\cos }^3{\theta }_r}\left({\sin }^4{\theta }_i-2n^2{\sin }^2{\theta }_i+n^2\right)$

The value of ${\theta }_i$ for maximum value of $b$ can be calculated as,

  $\begin{array}{l}\frac{db}{d{\theta }_i}=0\\ {\sin }^4{\theta }_i-2n^2{\sin }^2{\theta }_i+n^2=0\\ {\theta }_i={\sin }^{-1}\left(n\sqrt{1\sqrt{1-\frac{1}{n^2}}}\right)\end{array}$

Conclusion:

Therefore, the angle of incidence is, ${\sin }^{-1}\left(n\sqrt{1\sqrt{1-\frac{1}{n^2}}}\right)$ .

(b)

To determine

The angle of incidence.

Answer to Problem 38P

The angle of incidence is $48.5°$

Explanation of Solution

Given:

The index of refraction of glass is $n=1.60$ .

Formula used:

The expression for the angle of incidence is,

  ${\theta }_i={\sin }^{-1}\left(n\sqrt{1\sqrt{1-\frac{1}{n^2}}}\right)$

Calculation:

The angle of incidence can be calculated as,

  $\begin{array}{c}{\theta }_i={\sin }^{-1}\left(n\sqrt{1\sqrt{1-\frac{1}{n^2}}}\right)\\ ={\sin }^{-1}\left(1.6\times \sqrt{1\sqrt{1-\frac{1}{{\left(1.6\right)}^2}}}\right)\\ =48.5°\end{array}$

Conclusion:

Therefore, the angle of incidence is $48.5°$

(c)

To determine

The separation of the two beam.

Answer to Problem 38P

The separation of the two beam is $2.81\text{cm}$ .

Explanation of Solution

Given:

The thickness of the glass plate is $d=4\text{cm}$

Formula used:

The expression for angle of refraction can be given as,

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

The expression for $b$ can be given as,

  $b=2d\tan {\theta }_r\cos {\theta }_i$

Calculation:

The angle of refraction can be calculated as,

  $\begin{array}{c}{\theta }_r={\sin }^{-1}\left(\frac{n_1}{n_2}\sin {\theta }_i\right)\\ ={\sin }^{-1}\left(\frac{1}{1.6}\sin \left(48.5°\right)\right)\\ =27.9°\end{array}$

The separation of two beam can be calculated as,

  $\begin{array}{c}b=2d\tan {\theta }_r\cos {\theta }_i\\ =2\times \left(4\text{cm}\right)\tan \left(27.9°\right)\cos \left(48.5°\right)\\ =2.81\text{cm}\end{array}$

Conclusion:

Therefore, the separation of the two beam is $2.81\text{cm}$ .