Question 391: Ray Optics

Question

A monochromatic beam of light is incident at 60$$^\circ$$ on one face of an equilateral prism of refractive index n and emerges from the opposite face making an angle $$	heta$$(n) with the normal (see figure). For n = $$\sqrt 3 $$ the value of $$	heta$$ is 60$$^\circ$$ and $${{d	heta } \over {dn}} = m$$. The value of m is

Accepted Answer

Using Snell's law:
$$\sin 60^\circ  = n\sin {r_1}$$ ..... (1)
$$\sin {r_1} = {{\sqrt 3 } \over {2 	imes \sqrt 3 }} = {1 \over 2}$$
$${r_1} = 30^\circ $$
Also,
$$n\sin {r_2} = 1\sin 	heta $$
Also
$${r_1} + {r_2} = A = 60^\circ $$
Therefore,
$$n\sin (60^\circ  - {r_1}) = 1\sin 	heta $$ ....... (2)
Differentiating on both sides, we get
$$\sin (60^\circ  - {r_1}) - n\cos (60^\circ  - {r_1}){{d{r_1}} \over {dn}} = \cos 	heta {{d	heta } \over {dn}}$$
Differentiating Eq. (1) on both sides, we get
$$0 = \sin {r_1} + n\cos {r_1}{{d{r_1}} \over {dn}}$$
$$0 = {1 \over 2} + \sqrt 3 .{{\sqrt 3 } \over 2}{{d{r_1}} \over {dn}}$$
Therefore,
$${{d{r_1}} \over {dn}} = {{ - 1} \over 3}$$
Hence, substituting $${r_1} = 30^\circ $$, we get
$${{d{r_1}} \over {dn}} = {{ - 1} \over 3}$$
Now, $$\sin 30^\circ  - \sqrt 3 \cos 30^\circ \left( { - {1 \over 3}} ight) = \cos 60^\circ {{d	heta } \over {dn}}$$
$${1 \over 2} + {3 \over {2 	imes 3}} = {1 \over 2}{{d	heta } \over {dn}}$$
$${{d	heta } \over {dn}} = 2$$

Ray Optics

Integer / Numeric Answer Type