Question

The angle of deviation ($\delta$) vs angle of incidence (i) is plotted for a prism. Pick up the correct statements.

• A. The angle of prism is ${60}^0$
• B. The refractive index of the prism is $n=\sqrt{3}$
• C. For deviation to be ${65}^0$ the angle of incidence $i_1={55}^0$
• D. The curve of ${}^{\prime }{\delta }^{\prime }$ vs ${}^{\prime }{i}^{\prime }$ is parabolic

Answer 1

The correct option is C For deviation to be ${65}^0$ the angle of incidence $i_1={55}^0$

We know,

angle of deviation in a prism is given by,

$\delta =i+e-A$

and for minimum deviation, $i=e$

So minimum deviation, ${\delta }_m=2i-A$

From the graph, minimum deviation, ${\delta }_m={60}^{\circ }$, and the corresponding angle of incidence is $i={60}^{\circ }$

So $60=2\times 60-A⟹A={60}^{\circ }$

So for the given graph, the angle of the prism is ${60}^{\circ }$

Now, Refractive index of the prism is given by,

$n=\frac{\sin \left(\frac{A+{\delta }_m}{2}\right)}{\sin \frac{A}{2}}⟹n=\frac{\sin \left(\frac{{60}^{\circ }+{60}^{\circ }}{2}\right)}{\sin \frac{{60}^{\circ }}{2}}⟹n=\frac{\sin \left({60}^{\circ }\right)}{\sin {30}^{\circ }}⟹n=\frac{\frac{\sqrt{3}}{2}}{\frac{1}{2}}⟹n=\sqrt{3}$

So for the given graph the refractive index of the prism is $\sqrt{3}$

And for deviation, ${\delta }_1={65}^{\circ }$, we get

${\delta }_1=i_1+e-A$

substituting the corresponding values from the graph, we get

${65}^{\circ }=i_1+{70}^{\circ }-{60}^{\circ }⟹i_1={55}^{\circ }$

The curve of ′$\delta$′ vs ′$i$′ is not parabolic