Question

The angle of minimum deviation of a prism will be equal to its angle of refractive index. If refractive index of prism is:

• A. Between $\sqrt{2}$ and $2$
• B. Less than $1$
• C. Greater than $2$
• D. Between $\sqrt{2}$ and $1$

Answer 1

Answer: (A)

Between $\sqrt{2}$ and $2$

$n=\frac{sin\frac{A+{\delta }_m}{2}}{sin\frac{A}{2}}$
Given,
${\delta }_m=A$
$\therefore n=\frac{sin\frac{A+A}{2}}{sin\frac{A}{2}}=\frac{sinA}{sin\frac{A}{2}}$
$=\frac{2sin\frac{A}{2}cos\frac{A}{2}}{sin\frac{A}{2}}$
or $n=2cos\frac{A}{2}$
(i) When $A=0,$ then $cos\frac{A}{2}=1$
$\therefore n=2$
(ii) When $A={90}^{\circ },$ then
$cos\frac{A}{2}=cos\frac{{90}^{\circ }}{2}$
$=cos{45}^{\circ }=\frac{1}{\sqrt{2}}$
$\therefore n=2\times \frac{1}{\sqrt{2}}=\sqrt{2}$
$\therefore$ Value of $\mu$ lies between $\sqrt{2}$ and $2.$