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 $n _{2}\left( n _{2} \geq n _{1}\right)$ 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
$dn _{2} \sin 30^{\circ}=\cos\, e \,de$
or $dn _{2}=\Delta n\,\, de =\Delta e$
$\Delta n \sin 30^{\circ}=\cos e \Delta e$
$\Delta n \frac{1}{2}=\frac{\sqrt{7}}{4} \Delta e$
$\Delta n =\frac{\sqrt{7}}{2} \Delta e \ldots$ (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.