Question

what should be the answer for this?
give explanation.​

Answer 1

Answer:

d

Use Snell's law

$1. \sin( \theta) = \mu. \sin(r _1) \\ \\ \sin(r _1) = \frac{ \sin( \theta) }{ \mu} - - - - eq1$

Again for surface 2

$\mu \sin(r _2) = 1. \sin( \phi)$

Note that

$r _1 + r _2 = \frac{\pi}{2} \implies r _2 = \frac{\pi}{2} - r _1$

So

$\mu \sin( \frac{\pi}{2} - r _2) = \sin( \phi) \\ \\ \cos(r _1) = \frac{1}{ \mu} \sin( \phi) - - - - eq2$

Square and add eq1 and eq2

$\sin {}^{2} (r _1) + \cos {}^{2} (r _1) = \frac{ \sin {}^{2} ( \phi) }{ \mu {}^{2} } + {\sin {}^{2} ( \theta ) \over \mu {}^{2} } \\ \\ \sin {}^{2} ( \theta ) + \sin {}^{2} ( \mu) = { \mu}^{2}$