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(A) The light from the point source at the bottom of the tank will emerge from the surface only if the angle of incidence is less than or equal to the

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$2.6$

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(A) The light from the point source at the bottom of the tank will emerge from the surface only if the angle of incidence is less than or equal to the critical angle $C$.The radius $r$ of the circular area on the water surface is given by $r = h 	an C$, where $h = 80 \, cm = 0.8 \, m$.Since $\sin C = \frac{1}{\mu}$, we have $	an C = \frac{1}{\sqrt{\mu^2 - 1}}$.Thus, $r = \frac{h}{\sqrt{\mu^2 - 1}}$.Given $\mu = 1.33 \approx \frac{4}{3}$, we calculate $r = \frac{0.8}{\sqrt{(\frac{4}{3})^2 - 1}} = \frac{0.8}{\sqrt{\frac{16}{9} - 1}} = \frac{0.8}{\sqrt{\frac{7}{9}}} = \frac{0.8 	imes 3}{\sqrt{7}} = \frac{2.4}{2.646} \approx 0.907 \, m$.The area of the surface is $A = \pi r^2 = \pi 	imes (0.907)^2 \approx 3.14 	imes 0.8226 \approx 2.58 \, m^2$.Rounding to one decimal place, the area is approximately $2.6 \, m^2$.

$A$ small bulb is placed at the bottom of a tank containing water to a depth of $80 \, cm$. What is the area of the surface of water through which light from the bulb can emerge out? The refractive index of water is $1.33$. (Consider the bulb to be a point source.)

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