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Question: A point source of light is placed at a distance \(h\) below the surface of a large and deep lake. Wh...

A point source of light is placed at a distance hh below the surface of a large and deep lake. What fraction of light will escape through the surface of the water?

Explanation

Solution

We are given a light source placed at a certain distance below the surface of a large and deep-sea and we need to find the fraction of light escaping the surface of the lake that is light that can be perceived from outside the lake which will be due to the refraction of light rays depending upon their angle of incidence.

Complete step by step answer:
When the light source placed at the distance hh below the surface of the lake, the light rays coming from the source will be incident at the interface of the lake and air and if the angle of incidence is less than the critical angle, the light rays will be refracted and will escape through the surface of the water. Let the critical angle be θC{\theta _C} and we know that
θc=sin⁡−1(1μ){\theta _c} = {\sin ^{ - 1}}\left( {\dfrac{1}{\mu }} \right) where μ\mu is the refractive index of the denser with respect to the rarer medium.

Now to calculate the required fraction let us consider that the interface at which the refraction occurs is a part of a sphere of radius RR. So the solid angle subtended by the solid cone made up of the light rays from the source with an apex angle of 2θ2\theta is given by the formula;
Ω=2π(1−cos⁡θc)\Omega = 2\pi (1 - \cos {\theta _c})
Substituting the value of the critical angle we get
Ω=2π(1−1−(1μ)2)\Omega = 2\pi (1 - \sqrt {1 - {{\left( {\dfrac{1}{\mu }} \right)}^2}} )
Since cos⁡θc=1−(1μ)2\cos {\theta _c} = \sqrt {1 - {{\left( {\dfrac{1}{\mu }} \right)}^2}}
So the fraction of light energy lost can be expressed as the ratio of the solid angle subtended by the boundary to the solid angle subtended by the sphere which is the interface of water and air.
∴f=2π(1−1−(1μ)2)4π\therefore f = \dfrac{{2\pi (1 - \sqrt {1 - {{\left( {\dfrac{1}{\mu }} \right)}^2}} )}}{{4\pi }}

Note: We know a phenomenon called total internal reflection in which light rays entering from a denser medium to a rarer medium undergoes internal reflection when the angle of incidence of the light rays a the interface of the media is greater than the critical angle of the denser material. And all the light rays making the angle less than the critical angle will escape from the surface of the water. The fraction of light escaping out doesn't depend on the height at which the light source is kept.