A beam of light travelling in water is incident on a glass which is immersed in the water. The incident beam makes an angle of \(40^o\) with the normal. Calculate the angle of refraction in the glass.
[Refractive index of water = 1.33, Refractive index of glass = 1.5]
\(29.36^o\)
\(25.37^o\)
\(37.21^o\)
\(34.75^o\)
Correct answer is D
\(n_1\)=1.33; \(n_2\)=1.5; i=40°; r= ?
from snell's law:
\(n_1\) x sin i=\(n_2\)x sin r
⇒1.33 x sin 40°=1.5 x sin r
⇒sin r = 0.5700
⇒r = \(sin^{-1}\) (0.5700)
⇒ r = 34.75