QJEEAdv15Phe48e: JEE Advanced 2015 Physics - Class 12 JEEAdv15Phe48e Solution: A monochromatic beam of light is incident at 60 on one face of…

Using Snell's law:

$\sin 60^\circ = n\sin {r_1}$ ..... (1)

$\sin {r_1} = {{\sqrt 3 } \over {2 \times \sqrt 3 }} = {1 \over 2}$

${r_1} = 30^\circ$

Also,

$n\sin {r_2} = 1\sin \theta$

Also

${r_1} + {r_2} = A = 60^\circ$

Therefore,

$n\sin (60^\circ - {r_1}) = 1\sin \theta$ ....... (2)

Differentiating on both sides, we get

$\sin (60^\circ - {r_1}) - n\cos (60^\circ - {r_1}){{d{r_1}} \over {dn}} = \cos \theta {{d\theta } \over {dn}}$

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, $\sin 30^\circ - \sqrt 3 \cos 30^\circ \left( { - {1 \over 3}} \right) = \cos 60^\circ {{d\theta } \over {dn}}$

${1 \over 2} + {3 \over {2 \times 3}} = {1 \over 2}{{d\theta } \over {dn}}$

${{d\theta } \over {dn}} = 2$