1.

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

Answer»

`(4//3)SINI`
`1//sini`
`4//3`
1

Solution :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)`


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