Question

For a prism of prism angle $\theta ={60}^{\circ },$ the refractive indices of the left half and the right half are, respectively, $n_1\text{and}{n}_2(n_2\ge n_1)$ as shown in the figure. The angle of incidence $i$ is chosen such that the incident light rays will have minimum deviation if $n_1=n_2=n=\mathrm{1.5.}$ For the case of unequal refractive indices, $n_1=n\text{and}{n}_2=n+\Delta n$ $(where,\Delta n<<n)$, 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)

$\Delta e$ lies between $2.0$ and $3.0$ milliradians, if $\Delta n=2.8\times {10}^{-3}$.


Applying snell's law at the left surface,

$1\times \sin i=n_1\times \sin \left(\frac{\theta }{2}\right)$

$\Rightarrow 1\times \sin i=1.5\times \sin \left(\frac{{60}^{\circ }}{2}\right)$

$\Rightarrow \sin i=\frac{3}{4}$

Applying Snell's law, at the right surface,

$n_2\sin \frac{{60}^{\circ }}{2}=1\times \sin \left(e\right)$

$\Rightarrow 1.5\times \frac{1}{2}=\sin \left(e\right)(\because n_1=n_2=1.5)$

$\therefore \sin \left(e\right)=\frac{3}{4}\Rightarrow \cos \left(e\right)=\frac{\sqrt{7}}{4}$

Again, applying Snell's law, at the right surface,

$n_2\sin {30}^{\circ }=1\times \sin \left(e\right)$

On differentiating both sides, we get

$\left(\Delta n\right)\sin {30}^{\circ }=\cos \left(e\right)\times \left(\Delta e\right)(\because n_2=n+\Delta n;e=i+\Delta e)$

$\left(\Delta n\right)\times \frac{1}{2}=\frac{\sqrt{7}}{4}\times \left(\Delta e\right)$

$\therefore \Delta n>\Delta e$

Now, $\Delta e=\frac{2.8\times {10}^{-3}\times 2}{\sqrt{7}}=2.11\text{mrad}$

Hence, $\left(B\right)$ and $\left(C\right)$ are the correct answers.