There is a small source of light at some depth below the surface of water (refractive index $$= \frac{4}{3}$$) in a tank of large cross sectional surface area. Neglecting any reflection from the bottom and 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 $$2\pi rh$$]

Let the point source of light be at a distance $$r$$ from the water-air interface. Because the source emits light isotropically, the radiant energy is spread uniformly over the surface of a sphere of radius $$r$$, whose total area is $$4\pi r^2$$.

For a ray inside water to emerge into air it must strike the water surface with an angle of incidence $$\theta_i$$ that is less than or equal to the critical angle $$\theta_c$$. Using Snell’s law, $$n_1\sin\theta_i = n_2\sin\theta_t$$, and putting $$n_1 = \dfrac{4}{3}$$ (water), $$n_2 = 1$$ (air), and $$\theta_t = 90^\circ$$ for the limiting case, we have

$$\sin\theta_c = \frac{n_2}{n_1} = \frac{1}{\dfrac{4}{3}} = \frac{3}{4}.$$

Hence

$$\theta_c = \sin^{-1}\!\left(\frac{3}{4}\right)\approx 48.59^\circ.$$

We also need $$\cos\theta_c$$ because it will appear in the cap-area relation. Using $$\sin^2\theta_c + \cos^2\theta_c = 1$$, we get

$$\cos\theta_c = \sqrt{1-\sin^2\theta_c} = \sqrt{1-\left(\frac{3}{4}\right)^2} = \sqrt{1-\frac{9}{16}} = \sqrt{\frac{7}{16}} = \frac{\sqrt{7}}{4}\approx 0.6614.$$

Inside the spherical wavefront of radius $$r$$, all the rays that satisfy $$\theta_i \le \theta_c$$ form a cone of half-angle $$\theta_c$$. On the spherical surface this cone subtends a spherical cap. The formula for the area of a spherical cap of height $$h$$ carved out of a sphere of radius $$r$$ is

$$A_{\text{cap}} = 2\pi r h.$$

The height $$h$$ of the cap measured from the base circle to the top of the sphere relates to $$\theta_c$$ through simple geometry: the cap height is the difference between the sphere’s radius and the vertical projection of the radius that makes angle $$\theta_c$$ with the vertical. Thus

$$h = r(1-\cos\theta_c).$$

Substituting this $$h$$ in the cap-area formula gives

$$A_{\text{cap}} = 2\pi r \bigl[r(1-\cos\theta_c)\bigr] = 2\pi r^2(1-\cos\theta_c).$$

The fraction of the total spherical surface through which light may pass is therefore

$$\text{Fraction} = \frac{A_{\text{cap}}}{4\pi r^2} = \frac{2\pi r^2(1-\cos\theta_c)}{4\pi r^2} = \frac{1-\cos\theta_c}{2}.$$

Putting the numerical value $$\cos\theta_c \approx 0.6614$$ obtained earlier, we have

$$\text{Fraction} = \frac{1-0.6614}{2} = \frac{0.3386}{2} \approx 0.1693.$$

Multiplying by $$100$$ to convert this fraction into a percentage of the total emitted light,

$$\text{Percentage emerging} = 0.1693 \times 100 \approx 16.93\% \approx 17\%.$$

Hence, the correct answer is Option C.