A ray of light is incident at the glass-water interface at an angle i. It merges finally parallel to the surface of water. Then, the value of `mu_(g)` would be

Options

A. `(4//3)sini`
B. `1//sini`
C. `4//3`
D. 1

Answer

b.

Explanation

Applying Snell's law at glass-water surface,
`._(g)^(omega)mu=(sinr)/(sini)=(._(g)mu)/(._(omega)mu)` (i)
Applying Snell's law at water-air surface,
`._(g)^(omega)mu=(sin90^(@))/(sinr)=(._(g)mu)/(._(omega)mu)`
From (i) and (ii),
`(._(g)mu)/(._(omega)mu)=(3)/(4sini)rArr (3xx_(g)mu)/(4sini)rArr _gmu=(1)/(sini)`