Question

The angle of deviation $\left(\delta \right)$ vs angle of incidence $\left(i\right)$ is plotted for a prism. Pick up the incorrect statement

• A. The angle of prism is ${60}^{\circ }$.
• B. The refractive index of the prism is $\mu =\sqrt{3}$.
• C. For deviation to be ${65}^{\circ }$ the angle of incidence $i_1={55}^{\circ }$.
• D. The curve of $\delta$ vs $i$ is parabolic.

Answer 1

The correct option is D The curve of $\delta$ vs $i$ is parabolic.

Deviation produced by prism,
$\delta =(i+e)-A$
For minimum deviation, $(i=e)$
${\delta }_m=2i-A$
${60}^{\circ }=2\times {60}^{\circ }-A$
Thus, $A={60}^{\circ }$
$\therefore {60}^{\circ }$ is angle of prism.
$\Rightarrow {\delta }_m=i$
From the graph this condition matches for
$i={60}^{\circ }$

Now, refractive index at $i=A$ i.e. ${\delta }_m={60}^{\circ }$
$\mu =\frac{\sin \left(\frac{A+{\delta }_m}{2}\right)}{\sin \left(\frac{A}{2}\right)}$
$\Rightarrow \mu =\frac{\sin \left(\frac{{60}^{\circ }+{60}^{\circ }}{2}\right)}{\sin \left(\frac{{60}^{\circ }}{2}\right)}$
$\therefore \mu =\frac{\sqrt{3}}{2}\times 2=\sqrt{3}$
for $\delta ={65}^{\circ }$
$\Rightarrow {65}^{\circ }=i+e-A$
or, ${65}^{\circ }=i_1+{70}^{\circ }-{60}^{\circ }$ (from graph $e={70}^{\circ }$)
$\therefore i_1={55}^{\circ }$
The graph for $\delta vsi$ is not symmetrical and hence it doesn't follow parabolic curve.
$\therefore$ option (d) is correct.

Why this question? Caution: Always remember that the curve $\delta \text{vs}i$ is not parabolic