Question

The refracting angle of prism is A and refractive index of material of prism is $\cot \frac{A}{2}$ The angle of minimum deviation is

• A. $180^\circ -3 A$
• B. $180^\circ +2 A$
• C. $90^\circ - A$
• D. $180^\circ -2 A$

Answer 1

The Correct Option is D

For a prism, the relation between the refracting angle \( A \), the refractive index \( n \), and the angle of minimum deviation \( \delta \) is given by: \[ n = \cot \frac{A}{2} \] The angle of minimum deviation \( \delta \) is related to the refracting angle \( A \) by the equation: \[ \delta = 180^\circ - 2A \] Thus, the angle of minimum deviation is \( 180^\circ - 2A \).

Answer 2

$n=\frac{\sin \left(\frac{A+d_{m}}{2}\right)}{\sin \left(\frac{A}{2}\right)}$
$\Rightarrow \cot \left(\frac{A}{2}\right)=\frac{\sin \left(\frac{A+d_{m}}{2}\right)}{\sin \left(\frac{A}{2}\right)}$
$\Rightarrow \frac{\cos \left(\frac{A}{2}\right)}{\sin \left(\frac{A}{2}\right)}=\frac{\sin \left(\frac{A+d_{m}}{2}\right)}{\sin \left(\frac{A}{2}\right)}$
$\Rightarrow \sin \left(90-\frac{A}{2}\right)=\sin \left(\frac{A+d_{m}}{2}\right)$
$\Rightarrow 90-\frac{A}{2}=\frac{A+d_{m}}{2}$
$\Rightarrow 180-2 A=d_{m}$