Chapter 5, Problem 25CT

To determine

To calculate: The angle $\theta$ for the glass prism given that $n=\frac{\sin \left(\frac{\theta }{2}+\frac{\alpha }{2}\right)}{\sin \frac{\theta }{2}}$ .

Answer to Problem 25CT

The required angle is $76.52°$ .

Explanation of Solution

Given information:

The figure of the prism:

  

  $n=1.5$

  $\alpha =60°$

Formula used:

Trigonometric formula:

  $\cos 3\theta =4{\cos }^3\theta -3\cos \theta$

Calculation:

The given expression is $n=\frac{\sin \left(\frac{\theta }{2}+\frac{\alpha }{2}\right)}{\sin \frac{\theta }{2}}$ .

Simplify the right hand side of the above expression using the formula $\sin \left(x+y\right)=\sin x\cos y+\cos x\sin y$ .

  $\begin{array}{c}\frac{\sin \left(\frac{\theta }{2}+\frac{\alpha }{2}\right)}{\sin \frac{\theta }{2}}=\frac{\sin \frac{\theta }{2}\cos \frac{\alpha }{2}+\cos \frac{\theta }{2}\sin \frac{\alpha }{2}}{\sin \frac{\theta }{2}}\\ =\frac{\sin \frac{\theta }{2}\cos \frac{\alpha }{2}}{\sin \frac{\theta }{2}}+\frac{\cos \frac{\theta }{2}\sin \frac{\alpha }{2}}{\sin \frac{\theta }{2}}\\ =\cos \frac{\alpha }{2}+\cot \frac{\theta }{2}\sin \frac{\alpha }{2}\end{array}$

Substitute 1.5 for $\alpha$ in the above expression.

  $\begin{array}{c}\cos \frac{60°}{2}+\cot \frac{\theta }{2}\sin \frac{60°}{2}=\frac{\sqrt{3}}{2}+\frac{1}{\tan \frac{\theta }{2}}\frac{1}{2}\\ \tan \frac{\theta }{2}=\frac{1}{3-\sqrt{3}}\\ \frac{\theta }{2}=\arctan \left(\frac{1}{3-\sqrt{3}}\right)\\ \frac{\theta }{2}=38.26°\end{array}$

Solve for $\theta$ .

  $\begin{array}{c}\frac{\theta }{2}=38.26°\\ \theta =38.26°\times 2\\ =76.52°\end{array}$

Hence, the required angle is $76.52°$ .