Question

Angle of minimum deviation for a prism of refractive index 1.5 is equal to the angle of prism. The angle of prism is $(cos{41}^{\circ }=0.75)$

• A. ${62}^{\circ }$
• B. ${41}^{\circ }$
• C. ${82}^{\circ }$
• D. ${31}^{\circ }$

Answer 1

The correct option is C ${82}^{\circ }$

Given ${\delta }_m=A$, then by using $\mu =\frac{sin\frac{A+{\delta }_m}{2}}{sin\frac{A}{2}}\Rightarrow \mu =\frac{sin\frac{A+A}{2}}{sin\frac{A}{2}}=\frac{sinA}{sin\frac{A}{2}}=2cos\frac{A}{2}$
$\{sinA=2sin\frac{A}{2}cos\frac{A}{2}\}$
$\Rightarrow 1.5=2cos\frac{A}{2}\Rightarrow 0.75=cos\frac{A}{2}\Rightarrow {41}^{\circ }=\frac{A}{2}\Rightarrow A={82}^{\circ }$