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. $2.12\times 10^{8}m{s}^{-1}$
• B. $1.12\times 10^{8}m{s}^{-1}$
• C. $4.12\times 10^{8}m{s}^{-1}$
• D. $5.12\times 10^{8}m{s}^{-1}$

Answer 1

Answer: (A)

Using, $\mu =\frac{sin\left(A+{\delta }_m\right)/2}{sinA/2}$
Here, $A=\frac{\pi }{3}=60^{\circ },{\delta }_m=\frac{\pi }{6}=30^{\circ },$
$c=3\times 10^{8}m{s}^{-1}$
$\therefore \mu =\frac{sin\left(60^{\circ }+30^{\circ }\right)/2}{sin60^{\circ }/2}$
$=\frac{0.7071}{0.50}=1.414$
Therefore, $v=\frac{c}{\mu }=\frac{3\times 10^8}{10414}$
or $v=2.12\times 10^{8}m{s}^{-1}$