Question

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

• 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 $\left(\delta \right)$ vs 'i' is parabolic

Answer 1

Answer: (A), (B), (C)

The correct options are
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 $i_1={55}^{\circ }$

Angle of deviation in a prism is given by,
$\delta =i+e-A$
for minimum deviation $i=e$
$\therefore$ minimum deviation ${\delta }_m=2i-\text{A}$ ...(1)

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

So, using eq. (1) $60=2\times 60-A\because A={60}^{\circ }$
So for the given prism, the angle of the prism is ${60}^{\circ }$.

Now, the refractive index of the prism is given by,
$\mu =\frac{sin\left(\frac{a+{\delta }_m}{2}\right)}{sin\left(\frac{A}{2}\right)}\mu =\frac{sin\left(\frac{60+60}{2}\right)}{sin\left(\frac{60}{2}\right)}=\sqrt{3}$

So for the given graph the refractive index of the prism is $\sqrt{3}$ and for deviation, ${\delta }_1={65}^o$ , we get
${\delta }_1=i_1+e-A$
$65=i_1+70-60$
$i_1={55}^o$
Hence, the curve $\delta$ versus $i$ curve is not parabolic.

Hence, options A, B, and C are correct.