Light is incident from air on an oil layer at an incident angle of `30^(@)`. After moving through the oil 1, oil 2, and glass it enters water. If the refraction index of glass and water are `1.5` and `1.3`, respectively. Find the angle which the ray makes with the normal in water.

Explanation

As we know that `mu`sin i = constant
`rArrmu_(air) sini_((air))=mu_(glass)sinr_((glass))`
`sini_(glass)=(mu_(air))/(mu_(glass))sini_(air)`
Again `mu_(glass) sini_(glass)=mu_(water)sinr_(water)`
From Eqs. (i) and (ii), `sin 30^@ =1.3 sin r`
`rarr sinr= (1)/(2xx1.3)=(1)/(2.6),r=sin^(-1)(1/(2.6))`