A small bulb is placed at the bottom of a tank containing water to a depth of $$\sqrt{7}$$ m. The refractive index of water is $$\frac{4}{3}$$. The area of the surface of water through which light from the bulb can emerge out is $$x\pi$$ m$$^2$$. The value of $$x$$ is ______

We need to find the area of the water surface through which light from a bulb at the bottom can emerge.

Given that the depth of water is $$h = \sqrt{7}$$ m and the refractive index of water is $$\mu = \frac{4}{3}$$.

Since light can emerge from water only within the cone defined by the critical angle $$\theta_c$$, we have $$\sin \theta_c = \frac{1}{\mu} = \frac{3}{4}$$.

Now to determine the radius $$r$$ of the circle of light at the surface, note that $$\tan \theta_c = \frac{r}{h}$$. Using $$\sin \theta_c = \frac{3}{4}$$, we find $$\cos \theta_c = \sqrt{1 - \sin^2 \theta_c} = \sqrt{1 - \frac{9}{16}} = \sqrt{\frac{7}{16}} = \frac{\sqrt{7}}{4}$$, which gives $$\tan \theta_c = \frac{\sin \theta_c}{\cos \theta_c} = \frac{3/4}{\sqrt{7}/4} = \frac{3}{\sqrt{7}}$$. Substituting into $$r = h \times \tan \theta_c$$ yields $$r = \sqrt{7} \times \frac{3}{\sqrt{7}} = 3$$ m.

Next, the area is given by $$A = \pi r^2 = \pi \times 3^2 = 9\pi$$ m$$^2$$. Comparing this with $$x\pi$$, we obtain $$x = 9$$.

The value of $$x$$ is 9.