Question

Refractive index of the material of a prism is $1.5$. If ${\delta }_m=A$, what will be a value of angle of the given prism?
(where ${\delta }_m=$minimum deviation; A$=$ angle of prism).

• A. ${82.8}^o$
• B. ${41.4}^o$
• C. ${48.6}^o$
• D. ${90}^o$

Answer 1

The correct option is A ${82.8}^o$

Given, $\mu =1.5$ and ${\delta }_m=A$
$\mu =\frac{\sin \left(\frac{A+{\delta }_m}{2}\right)}{\sin \frac{A}{2}}$

$1.5=\frac{sin\left(\frac{A+A}{2}\right)}{\sin \frac{A}{2}}=\frac{\sin A}{\sin \frac{A}{2}}$

$=\frac{2\sin \frac{A}{2}\cos \frac{A}{2}}{\sin \frac{A}{2}}$

$1.5=2\cos \frac{A}{2}$

$\cos \frac{A}{2}=\frac{1.5}{2}=0.75=\cos 41.4$

$\frac{A}{2}=41.4$
$A=82.8$.