Question

Two rays of light are incident normally on a tub of height `H` filled with water. They pass through glass slabs of heights `h_1` and `h_2`. If the speed of light in vacuum is `c`, what is the time difference between the rays of light to reach the bottom?

Refractive indices of water and glass are `4/3` and `3/2` respectively.

Answer 1

The situation described in the problem is shown in the above figure
As the rays of light are incident normally , the rays will refracted through the media without deviation.

For slab of height `h_1`

the total path length of water and glass =H

Path length of glass medium`=h_1`

Then path length of water medium`=H-h_1`

For slab of height `h_2`

the total path length of water and glass =H

Path length of glass medium`=h_2`

Then path length of water medium`=H-h_2`

We know refractive index of water

`mu_w=4/3="velocity of light in vacuum"/("velocity of light in water"(v_w))`

`=>4/3=c/v_w=>v_w=(3c)/4`

Again we know refractive index of glass

`mu_g=3/2="velocity of light in vacuum"/("velocity of light in glass"(v_g))`

`=>3/2=c/v_g=>v_g=(2c)/3`

Calculation of time to reach `color(red)(ray_1)` at the bottom from surface

So time taken by the light `ray_1` to pass through water of path length `=H-h_1` is given by

`t_w=(H-h_1)/v_w=(4(H-h_1))/(3c)`

And time taken by the light `ray_1` to pass through glass of path length `h_1` is given by

`t_g=h_1/v_g=(3h_1)/(2c)`

Hence total time `t_1` to reach `color(red)(ray_1)` at the bottom from surface

`color(blue)(t_1=t_w+t_g)`

Calculation of time to reach `color(red)(ray_2)` at the bottom from surface

Now time taken by the light `ray_2` to pass through water of path length `=H-h_2` is given by

`t'_w=(H-h_2)/v_w=(4(H-h_2))/(3c)`

And time taken by the light `ray_2` to pass through glass of path length `h_2` is given by

`t'_g=h_2/v_g=(3h_2)/(2c)`

Hence total time `t_1` to reach `color(red)(ray_2)` at the bottom from surface

`color(blue)(t_2=t'_w+t'_g)`

Hence the time difference between the two rays of light to reach the bottom is given by

`Deltat=abs(t_1-t_2)=abs((t_w+t_g)-(t'w+t'g))`

`=abs((4(H-h_1))/(3c)+(3h_1)/(2c)-(4(H-h_2))/(3c)-(3h_2)/(2c))`

`=abs((8H-8h_1 +9h_1-8H+8h_2-9h_2)/(6c))`

`=abs((h_1-h_2)/(6c))`