A small bulb (assumed to be a point source) ...

CBSE Class 12 Physics 2013 Paper

Question : 20 of 29

Marks: +1, -0

A small bulb (assumed to be a point source) is placed at the bottom of a tank containing water to a depth of

80 \text{ cm}

. Find out the area of the surface of water through which light from the bulb can emerge. Take the value of the refractive index of water to be

\;\frac{4}{3}

Solution:

Using Snell's law,

\;\mu = \frac{\sin i}{\sin r}

\;\; \text{ Or, } \;\; \;\frac{4}{3} = \frac{\sin 90^{\circ}}{\sin r}

\; \therefore \;\; i = \sin^{-1} \frac{3}{4} = 48.59^{\circ}

\;\; \text{ Now, } \;\; \tan i = \frac{AB}{OB} = \frac{R}{0.8}

\; \therefore \;\; R = 0.8 \times \tan 48.59^{\circ}

\; = 0.8 \times 1.134 = 0.9 \text{ m}

Area of surface of water through which light will emerge

\; = \pi R^2 = \pi \times (0.9)^2

\; = 2.54 \text{ m}^2

Go to Question:

Prev Question

Next Question