Question

A ray in glass is incident onto a water-glass interface. The angle of incidence equals half the critical angle for that interface. The index of refraction for water is 1.33 and for glass is 1.78. What is the angle that the refracted ray in the water makes with the normal?

A- 43°
B- 33°
C- 38°
D- 28°
E- 23°

Answer 1

My answert would be A but i was told it was wrong

Answer 2

pls tell your solution i'll help correcting it if theres any mistake

Answer 3

hint:
use snell's law which says that \(\mu_1sini=\mu_2sinr\)
here \(\mu_1\)=refractive index of the medium of incident ray
\(\mu_2\)=refractive index of medium of refracted ray
\(i\)=angle of incidence
\(r\)=angle of refraction

critical angle is the angle at which if the light is incident then it doesn't gets refracted rather it gets reflected and it just goes off grazing the interface of the medium

like this-
here \(\theta_c\) is the critical angle
so the angle of refraction is 90 degress

using this info you can find the critical angle by applying snell's law
and half of critical angle is the angle of incidence (given in the question) once u get this angle then again apply snell's law to find angle of refraction