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

Answer 1

The Correct Option is A

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

Answer 2

For a prism, the relationship between the refractive index \( n \), the angle of deviation \( D \), and the refracting angle \( A \) is given by: \[ n = \sin \left( \frac{A + D}{2} \right) \div \sin \left( \frac{A}{2} \right) \] However, we are also given that the refractive index is \( \cot \frac{A}{2} \). So we can write: \[ \cot \left( \frac{A}{2} \right) = \frac{\sin \left( \frac{A + D}{2} \right)}{\sin \left( \frac{A}{2} \right)} \] At minimum deviation, the angle of deviation is \( D_{\text{min}} \), and the relationship is given by: \[ D_{\text{min}} = 180^\circ - A - \text{(this is a general result for minimum deviation in a prism)} \] Thus, the angle of minimum deviation is \( 180^\circ - 2A \), which corresponds to option \({A} \).