Chapter 31, Problem 83P

(a)

To determine

The expression for $\frac{d{\varphi }_{\text{d}}}{d{\theta }_1}$ .

Answer to Problem 83P

The expression for $\frac{d{\varphi }_{\text{d}}}{d{\theta }_1}$ is $2-\frac{4\cos {\theta }_1}{\sqrt{n^2-{\sin }^2{\theta }_1}}$ .

Explanation of Solution

Formula used:

The expression for ${\varphi }_{\text{d}}$ is given by,

  ${\varphi }_{\text{d}}=\pi +2{\theta }_1-4{\sin }^{-1}\left(\frac{n_{\text{air}}\sin {\theta }_1}{n_{\text{water}}}\right)$

Calculation:

The expression for ${\varphi }_{\text{d}}$ is calculated as,

  $\begin{array}{c}{\varphi }_{\text{d}}=\pi +2{\theta }_1-4{\sin }^{-1}\left(\frac{n_{\text{air}}\sin {\theta }_1}{n_{\text{water}}}\right)\\ =\pi +2{\theta }_1-4{\sin }^{-1}\left(\frac{\left(1.00\right)\sin {\theta }_1}{n}\right)\\ =\pi +2{\theta }_1-4{\sin }^{-1}\left(\frac{\sin {\theta }_1}{n}\right)\end{array}$ ............ (1)

Differentiate equation (1) with respect to ${\theta }_1$ .

  $\begin{array}{c}\frac{d{\varphi }_{\text{d}}}{d{\theta }_1}=\frac{d}{d{\theta }_1}\left(\pi +2{\theta }_1-4{\sin }^{-1}\left(\frac{\sin {\theta }_1}{n}\right)\right)\\ =2-\frac{4\cos {\theta }_1}{\sqrt{n^2-{\sin }^2{\theta }_1}}\end{array}$

Conclusion:

Therefore, the expression for $\frac{d{\varphi }_{\text{d}}}{d{\theta }_1}$ is $2-\frac{4\cos {\theta }_1}{\sqrt{n^2-{\sin }^2{\theta }_1}}$ .

(b)

To determine

The proof that $\cos {\theta }_{\text{lm}}=\sqrt{\frac{1}{3}\left(n^2-1\right)}$ .

Answer to Problem 83P

It is proved that $\cos {\theta }_{\text{lm}}=\sqrt{\frac{1}{3}\left(n^2-1\right)}$ .

Explanation of Solution

Calculation:

Consider the expression,

  $\frac{d{\varphi }_{\text{d}}}{d{\theta }_1}=2-\frac{4\cos {\theta }_1}{\sqrt{n^2-{\sin }^2{\theta }_1}}$ ............ (2)

Substitute $0$ for $\frac{d{\varphi }_{\text{d}}}{d{\theta }_1}$ in equation (2).

  $\begin{array}{c}0=2-\frac{4\cos {\theta }_1}{\sqrt{n^2-{\sin }^2{\theta }_1}}\\ 16{\cos }^2{\theta }_1=4\left(n^2-{\sin }^2{\theta }_1\right)\\ 12{\cos }^2{\theta }_{\text{lm}}=4n^2-4\left(\cos {\theta }_{\text{lm}}=\cos {\theta }_1\right)\\ \cos {\theta }_{\text{lm}}=\sqrt{\frac{n^2-1}{3}}\end{array}$

Conclusion:

Therefore, it is proved that $\cos {\theta }_{\text{lm}}=\sqrt{\frac{1}{3}\left(n^2-1\right)}$ .

(c)

To determine

The angular separation of red and blue light.

Answer to Problem 83P

The angular separation for blue light is $139.42°$ and for red light is $137.75°$ .

Explanation of Solution

Formula used:

The expression for angular separation is given by,

  ${\varphi }_{\text{d}}=\pi +2{\cos }^{-1}\left[\sqrt{\frac{n^2-1}{3}}\right]-4{\sin }^{-1}\left(\frac{\sin \left\{{\cos }^{-1}\left[\frac{n^2-1}{3}\right]\right\}}{n}\right)$

Calculation:

The angular separation for blue light is calculated as,

  $\begin{array}{c}{\varphi }_{\text{d,blue}}=\pi +2{\cos }^{-1}\left[\sqrt{\frac{n^2-1}{3}}\right]-4{\sin }^{-1}\left(\frac{\sin \left\{{\cos }^{-1}\left[\frac{n^2-1}{3}\right]\right\}}{n}\right)\\ =\pi +2{\cos }^{-1}\left[\sqrt{\frac{{\left(1.3435\right)}^2-1}{3}}\right]-4{\sin }^{-1}\left(\frac{\sin \left\{{\cos }^{-1}\left[\frac{{\left(1.3435\right)}^2-1}{3}\right]\right\}}{1.3435}\right)\\ =139.42°\end{array}$

The angular separation for red light is calculated as,

  $\begin{array}{c}{\varphi }_{\text{d,red}}=\pi +2{\cos }^{-1}\left[\sqrt{\frac{n^2-1}{3}}\right]-4{\sin }^{-1}\left(\frac{\sin \left\{{\cos }^{-1}\left[\frac{n^2-1}{3}\right]\right\}}{n}\right)\\ =\pi +2{\cos }^{-1}\left[\sqrt{\frac{{\left(1.3318\right)}^2-1}{3}}\right]-4{\sin }^{-1}\left(\frac{\sin \left\{{\cos }^{-1}\left[\frac{{\left(1.3318\right)}^2-1}{3}\right]\right\}}{1.3318}\right)\\ =137.75°\end{array}$

Conclusion:

Therefore, the angular separation for blue light is $139.42°$ and for red light is $137.75°$ .