Question

For a prism of prism angle $\theta =60^{\circ }$, the refractive indices of the left half and the right half are, respectively, $n_1$ and $n_2\left(n_2\ge n_1\right)$ as shown in the figure. The angle of incidence $i$ is chosen such that the incident light rays will have minimum deviation if $n_1=n_2=n=1.5$. For the case of unequal refractive indices, $n_1=n$ and $n_2=$ $n+\Delta n$ (where $\Delta n<<n$ ), the angle of emergence $e=i+\Delta e$. Which of the following statement(s) is(are) correct?

• A. The value of $\Delta e$ (in radians) is greater than that of $\Delta n$
• B. $\Delta e$ is proportional to $\Delta n$
• C. $\Delta e$ lies between $2.0$ and $3.0$ milliradians, if $\Delta n=2.8\times 10^{-3}$
• D. $\Delta e$ lies between $1.0$ and $1.6$ milliradians, if $\Delta n=2.8\times 10^{-3}$

Answer 1

Answer: (B), (C)

Diagram at minimum deviation for $n_1=n_2=n$
$n=1.5$
$r_1=r_2=\theta /2=30^{\circ }$
for face $AQ$
$n\sin r_2=\sin e$
$1.5\sin 30^{\circ }=\frac{3}{2}\times \frac{1}{2}=\sin e$
$\sin e=\frac{3}{4},\cos e=\frac{\sqrt{7}}{4}$
When $n_2$ is given small variation there will be no change in path of light ray inside prism.
As deviation on face $AC$ is zero.
So, $r_2=30^{\circ }$
Now for face $AQ$
$n_2\sin 30^{\circ }=\sin e$
for small change in $n_2$ change in $e$ is given by
$d{n}_2\sin 30^{\circ }=\cos ede$
or $d{n}_2=\Delta nde=\Delta e$
$\Delta n\sin 30^{\circ }=\cos e\Delta e$
$\Delta n\frac{1}{2}=\frac{\sqrt{7}}{4}\Delta e$
$\Delta n=\frac{\sqrt{7}}{2}\Delta e\dots$ (i)
$\Delta n>\Delta e$
$\Delta n\propto \Delta e$
Hence, option (B) is correct.
$\Delta e=\frac{2.8\times 10^{-3}\times 2}{\sqrt{7}}$
Hence, option (C) is correct.