Question

A light ray enters a glass enclosed fish tank. From air it enters the glass at 20.° with respect to the surface, then emerges into the water. The index for glass is 1.50 and for water 1.33.

(a) What is the angle of refraction in the glass?
(b) What is the angle of refraction in the water?
(c) Is there any incident angle in air for which the ray will not enter the water due to total internal reflection?

Answer 1

a)
`\angle r_(ag)=38.79^(\circ)`

b)
`\angle r_(gw)=44.95^(\circ)`

c) not possible

Step-by-step explanation:

Given:

angle of incidence on the air-glass interface,
`\angle i_(ag)=90-20=70^(\circ)`

refractive index of glass with respect to air,
`n_g=1.5`

refractive index of water with respect to air,
`n_a=1.33`

a)

For angle of refraction in glass we use Snell's law:

`n_g=\frac{sin\ i_(ag)}{sin\ r{ag}}`

`1.5=(sin\ 70)/(sin\ r_(ag))`

`\angle r_(ag)=38.79^(\circ)`

b)

Now we have angle of incident for glass-water interface,
`\angle i_(gw)=\angle r_(ag)=38.79^(\circ)`

And the refractive index of water with respect to glass:

`n_(gw)=(n_w)/(n_g)`

`n_(gw)=(1.33)/(1.5)`

`n_(gw)=0.8867`

Therefore, angle of refraction in the water:

`n_(gw)=(sin\ i_(gw))/(sin\ r_(gw))`

`0.8867 =(sin\ 38.79)/(sin\ r_(gw))`

`\angle r_(gw)=44.95^(\circ)`

c)

For total internal reflection through water the light must enter the glass-water interface at an angle greater than the critical angle.

So,

`n_(gw)=(sin\ i_((gw)_c))/(sin\ 90)`

`0.8867 =(sin\ i_((gw)_c))/(1)`

`i_((gw)_c)=62.46^(\circ)`

Now this angle will become angle of refraction for the air-glass interface.

Hence,

`n_g=(sin\ i_(ag))/(sin\ i_((gw)_c))`

`1.5=(sin\ i_(ag))/(0.8867 )`

`sin\ i_(ag)=1.33005`

Since we do not have any angle for sine for which the value exceeds 1, therefore it is not possible that the light will reflect from the water.