Question

If minimum deviation produced by a prism is equal to prism angle. then refractive index of material of prism can be

• A. 1.5
• B. 1.6
• C. 1.9
• D. 1.3

Answer 1

Answer: (A), (B), (C)

The correct options are
A 1.6
B 1.5
C 1.9

If$A$ is the prism angle and ${\delta }_m$ is the minimum angle deviation, then-

$\mu =\frac{\mathrm{Sin}\left(\frac{A+{\delta }_m}{2}\right)}{\mathrm{Sin}\left(\frac{A}{2}\right)}$

In the case when $A={\delta }_m$, then we have-

$\mu =\frac{\mathrm{Sin}\left(\frac{A+A}{2}\right)}{\mathrm{Sin}\left(\frac{A}{2}\right)}$

$\mu =\frac{\mathrm{Sin}\left(A\right)}{\mathrm{Sin}\left(\frac{A}{2}\right)}$

$\mu =\frac{2.\mathrm{Sin}\left(\frac{A}{2}\right).\mathrm{Cos}\left(\frac{A}{2}\right)}{\mathrm{Sin}\left(\frac{A}{2}\right)}$

$\mu =2.\mathrm{Cos}\left(\frac{A}{2}\right)$

Now, the range of $A$ is $0^{\circ }$ to ${90}^{\circ }$

When $A=0^{\circ }$,

$\mu =2.\mathrm{Cos}\left(0^{\circ }\right)=2$

When $A={90}^{\circ }$,

$\mu =2.\mathrm{Cos}\left({45}^{\circ }\right)=2\times \frac{1}{\sqrt{2}}=\sqrt{2}=1.414$

So, the range of $\mu$ is from $1.414$ to $2$. Thus, option (A), (B) & (C) are correct.