Question

In the diving pool in the Richards Building, there is a glass window under the water. There is a room full of air on the other side of the glass window pane. If I dive under the water and look at the window, I notice that I can see the room on the other side of the window only through a certain area. Outside that circle, light undergoes total internal reflection at the surface between the glass and the air, and the window appears like a mirror. If I am 1.01 m from the window, find the radius of the circle. The index of refraction of water is 1.33, and the index of refraction of glass is 1.52. Be sure to take into account that the light ray which is refracted at the glass-air surface is also refracted at the glass-water surface and changes direction there too.

Answer 1

`R = 1.15 m`

Step-by-step explanation:

For total internal reflection on the glass air interface we know that

`\mu_g sin\theta = \mu_(air) sin90`

`1.52 sin\theta = 1`

`\theta = 41.1 degree`

now this is the angle at glass air interface

so now for the refraction at water glass interface we will have

`\mu_w sin \phi = \mu_g sin\theta`

`1.33 sin\phi = 1.52 sin41.1`

`\phi = 48.75 degree`

now for the radius of the circle we know that

`tan\phi = (R)/(H)`

`tan48.75 = (R)/(1.01)`

`R = 1.15 m`