Question

A ray in glass is incident onto a water-glass interface, at an angle of incidence equal to half the critical angle for that interface. The indices of refraction for water and the glass are 1.33 and 1.43, respectively. The angle that the refracted ray in the water makes with the normal is closest to:

A) 37° B) 42° C) 47° D) 32° E) 27°

Answer 1

A) 37°

Step-by-step explanation:

The formula for the critical angle is:

`{sin\theta_(critical)}=\frac {n_r}{n_i}`

Where,

`{\theta_(critical)}` is the critical angle

`n_r` is the refractive index of the refractive medium. = 1.33

`n_i` is the refractive index of the incident medium. = 1.43

Applying in the formula as:

`{sin\theta_(critical)}=(1.33)/(1.43)`

The critical angle is = sin⁻¹ 0.93006 = 68.4442°

Angle of incidence is half of the critical angle. So,

Angle of incidence = 68.4442° / 2 = 34.2221°

Using Snell's law as:

`n_i* {sin\theta_i}={n_r}*{sin\theta_r}`

Where,

`{\theta_i}` is the angle of incidence ( 34.2221° )

`{\theta_r}` is the angle of refraction ( ? )

`{n_r}` is the refractive index of the refraction medium (water, n=1.33)

`{n_i}` is the refractive index of the incidence medium (glass, n=1.43)

Hence,

`1.43* {sin34.2221^0}={1.33}*{sin\theta_r}`

Angle of refraction= sin⁻¹ 0.6047 = 37°.