Question

There is a small source of light at some depth below the surface of water (refractive index $$=4/3$$) in a tank of large cross sectional surface area. Neglecting any reflection from the bottom and the absorption by water, percentage of light that emerges out of surface is (nearly):
[Use the fact that surface area of a spherical cap of height $$h$$ and radius of curvature $$r$$ is $$2\pi r h$$]

Answer 1

Correct option is C. $$17\%$$
$$\sin\theta=\dfrac{1}{\mu}$$ here angle is $$\beta$$

$$\Rightarrow \dfrac{1}{\mu}=\sin \beta=\dfrac{1}{4/3}$$

$$\Rightarrow \sin \beta=\dfrac{3}{4}$$

$$\Rightarrow \cos \beta=\dfrac{\sqrt{7}}{4}$$

We know

Solid angle, $$d\Omega=2\pi R^2(1-\cos\beta)$$

Percentage of light $$=\dfrac{2\pi R^2(1-\cos\beta)}{4\pi R^2}\times 100$$

$$=\dfrac{1-\cos\beta}{2}\times 100=\left(\dfrac{4-\sqrt{7}}{8}\right)\times 100$$

$$\boxed{=17\%}$$