Chapter 5, Problem 13PS

a.

To determine

Write the index of refraction as a function of $\cot \left(\frac{\theta }{2}\right)$ .

Answer to Problem 13PS

  $\boxed{\cos \frac{\alpha }{2}+\cot \frac{\theta }{2}\sin \frac{\alpha }{2}}$

Explanation of Solution

Given information:

The index of refraction n of a transparent material is the ratio of the speed of light in a vacuum to the speed of light in the material. Some common materials and their indices of refraction are air $\left(1.00\right)$ , water $\left(1.33\right)$ , and glass $\left(1.50\right)$ . Triangular prisms are often used to measure the index of refraction based on the formula

  $n=\frac{\sin \left(\frac{\theta }{2}+\frac{\alpha }{2}\right)}{\sin \frac{\theta }{2}}$

For the prism shown in the figure, $\alpha ={60}^{\circ }$

Calculation:

Let us consider that the relation between the index of refraction and angle of deviation is given as below:

  $n=\frac{\sin \left(\frac{\theta }{2}+\frac{\alpha }{2}\right)}{\sin \frac{\theta }{2}}$

Also consider the following figure of prism,

  

To obtain the above relation in terms of $\cot \frac{\theta }{2}$ , we need to approach the result as follows:

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

Hence, the index of refraction as a function of $\cot \left(\frac{\theta }{2}\right)$ is $\boxed{\cos \frac{\alpha }{2}+\cot \frac{\theta }{2}\sin \frac{\alpha }{2}}$

b.

To determine

Find $\theta$ for a prism made of glass.

Answer to Problem 13PS

  $\boxed{\theta ={76.128}^{\circ }}$

Explanation of Solution

Given information:

The index of refraction n of a transparent material is the ratio of the speed of light in a vacuum to the speed of light in the material. Some common materials and their indices of refraction are air $\left(1.00\right)$ , water $\left(1.33\right)$ , and glass $\left(1.50\right)$ . Triangular prisms are often used to measure the index of refraction based on the formula

  $n=\frac{\sin \left(\frac{\theta }{2}+\frac{\alpha }{2}\right)}{\sin \frac{\theta }{2}}$

For the prism shown in the figure, $\alpha ={60}^{\circ }$

Calculation:

Let us consider that the relation between the index of refraction and angle of deviation is given as below:

  $n=\frac{\sin \left(\frac{\theta }{2}+\frac{\alpha }{2}\right)}{\sin \frac{\theta }{2}}$

Also consider the following figure of prism,

  

The index of refraction of glass is equal to $1.5$ and the minimum angle of deviation $\alpha$ is ${60}^{\circ }$ for given prism.

From the given information, we can calculate the angle of deviation for th given prism which is by above formula,

  $\begin{array}{l}1.5=\cos \left(\frac{{60}^{\circ }}{2}\right)+\cot \frac{\theta }{2}\sin \left(\frac{{60}^{\circ }}{2}\right)\\ 1.5-\cos {30}^{\circ }=\cot \frac{\theta }{2}\sin {30}^{\circ }\\ \cot \frac{\theta }{2}=\frac{1.5-\frac{\sqrt{3}}{2}}{\frac{1}{2}}\\ \cot \frac{\theta }{2}=3-\sqrt{3}\\ \frac{\theta }{2}={\cot }^{-1}\left(3-\sqrt{3}\right)\end{array}$

  $\begin{array}{l}\theta =2{\cot }^{-1}\left(3-\sqrt{3}\right)\\ \theta =2\times ({38.064}^{\circ })\end{array}$

  $\boxed{\theta ={76.128}^{\circ }}$

Hence, $\boxed{\theta ={76.128}^{\circ }}$