Question

The angle of minimum deviation for prism of angle $\pi/3$ is $\pi/6$, if the velocity of light in vacuum is $3 \times 10^8 \,m s^{-1}$, then the velocity of light in material of the prism is

• A. $2.12\times 10^8 \,ms^{-1}$
• B. $1.12\times 10^8 \,ms^{-1}$
• C. $4.12\times 10^8 \,ms^{-1}$
• D. $5.12\times 10^8 \,ms^{-1}$

Answer 1

Answer: (A)

Using, $\mu=\frac{sin \left(A +\delta_{m}\right)/2}{sin A/2} $
Here, $A = \frac{\pi}{3} = 60^{\circ}, \delta_{m} = \frac{\pi}{6} =30^{\circ},$
$c = 3 \times 10^{8} ms^{-1}$
$ \therefore \mu = \frac{sin\left(60^{\circ}+30^{\circ}\right)/2}{sin 60^{\circ}/2} $
$ = \frac{0.7071}{0.50} = 1.414$
Therefore, $v= \frac{c}{\mu} = \frac{3 \times10^{8}}{10414}$
or $v= 2.12 \times 10^{8} ms^{-1}$