Question

The refracting angle of a prism is $A$ and the refractive index of the material of the prism is $cot\left(\frac{A}{2}\right)$. The angle of minimum deviation is

• A. $180°-2A$
• B. $90°-2A$
• C. $180°+2A$
• D. $180°-3A$

Answer 1

The correct option is A

$180°-2A$

Step 1: Given parameters

Refracting angle of a prism is $A$.

The refractive index of the material of the prism; $\mu =cot\left(\frac{A}{2}\right)$.

Step 2: Formula used

Use the relation of refraction through a prism as follows:

$\mu =\frac{\sin {\displaystyle \frac{A+{\delta }_m}{2}}}{\sin {\displaystyle \frac{A}{2}}}$…………….(1)

Where, $A$ is the angle of the prism, ${\delta }_m$ is the angle of minimum deviation

$\mu =\frac{\cos {\displaystyle \frac{A}{2}}}{\sin {\displaystyle \frac{A}{2}}}$ (By rearranging)

Step 3: Calculation

Substitute the value of $\mu$in equation (1).

$cot\frac{A}{2}=\frac{\sin \left({\displaystyle \frac{{\delta }_m+A}{2}}\right)}{\sin \left({\displaystyle \frac{A}{2}}\right)}$

$\frac{\cos {\displaystyle \frac{A}{2}}}{\sin {\displaystyle \frac{A}{2}}}=\frac{\sin \left({\displaystyle \frac{{\delta }_m+A}{2}}\right)}{\sin \left({\displaystyle \frac{A}{2}}\right)}$

$\sin \left(\frac{\pi }{2}-\frac{A}{2}\right)=\sin \left(\frac{{\delta }_m+A}{2}\right)$

$\frac{\pi }{2}-\frac{A}{2}=\frac{{\delta }_m+A}{2}$

After simplification,

${\delta }_m=\left(180°-2A\right)$

Hence the angle of minimum deviation is $180°-2A$

Hence the correct option is (A)