Question

if angle of minimum deviation of a prism is equal to refracting angle of the prism, then refractive index $\mu$ of material of prism satisfies

Answer 1

Step 1: Given

The refractive index of the prism is $\mu$.

Let ${\delta }_m$ be the angle of minimum deviation and $A$ be the refracting angle of the prism.
Then as per given data, ${\delta }_m=A$

Step 2: Formulas used

We know that,
$\mu =\frac{\sin \left(A+{\displaystyle \frac{{\delta }_m}{2}}\right)}{\sin \left({\displaystyle \frac{A}{2}}\right)}$

Step 3: Calculate the refractive index

Given ${\delta }_m=A$, thus,

$\begin{array}{ccc}& \mu =\frac{\sin \left(A+{\displaystyle \frac{A}{2}}\right)}{\sin \left({\displaystyle \frac{A}{2}}\right)}& \\ \Rightarrow & \mu =\frac{\sin \left(A\right)\cos \left({\displaystyle \frac{A}{2}}\right)+\cos \left(A\right)\sin \left({\displaystyle \frac{A}{2}}\right)}{\sin \left({\displaystyle \frac{A}{2}}\right)}& \left[\because \sin \left(x+y\right)=\sin x\cos y+\cos x\sin y\right]\\ \Rightarrow & \mu =\frac{\sin A\cos \left({\displaystyle \frac{A}{2}}\right)}{\sin \left({\displaystyle \frac{A}{2}}\right)}+\frac{\cos A\sin \left({\displaystyle \frac{A}{2}}\right)}{\sin \left({\displaystyle \frac{A}{2}}\right)}& \\ \Rightarrow & \mu =\frac{2\sin \left({\displaystyle \frac{A}{2}}\right)\cos \left({\displaystyle \frac{A}{2}}\right)\cos \left({\displaystyle \frac{A}{2}}\right)}{\sin \left({\displaystyle \frac{A}{2}}\right)}+\cos A& \\ \Rightarrow & \mu =2{\cos }^2\left(\frac{A}{2}\right)+\cos A& \\ \Rightarrow & \mu =2\left[\frac{1+\cos A}{2}\right]+\cos A& \left[\because \cos \left(\frac{x}{2}\right)=\sqrt{\frac{1+\cos x}{2}}\right]\\ \therefore & \mu =1+2\cos A& \end{array}$

The angle of prism $A$ can only lie between $0°$ and $90°$.

Thus, $\mu$ lies between $\mu =1+2\cos 90°=1$ and $\mu =1+2\cos 0°=3$

Therefore, the range of the refractive index will be $1\le \mu \le 3$.