Question

Dez pours a liquid (n = 1.398) into a container made of glass (n =1.541 ). The light ray in glass incident on the glass-to-liquid boundary makes an angle of 43.2° with the normal. Find the angle of the corresponding refracted ray.

Answer 1

`\theta_(r)`= 48.98°≅49°

Step-by-step explanation:

refractive index of liquid n_{liquid}= 1.398

refractive index of glass n_{glass}= 1.541

angle of incidence i= 43.2°

we have to find corresponding angle of refraction

by snell's law we write

`n_(glass)sin\theta_i=n_(water)sin\theta_r`

now putting values

`1.541sin43.2=1.398sin\theta_r`

θ_{r}=
`sin^(-1)((1.541sin(43.2))/(1.398) )`

on calculating we get

`\theta_(r)`= 48.98°≅49°