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(D) Let $x$ be the height of the pole above the water surface.The angle of incidence $i$ with the normal is $90^{\circ} - 30^{\circ} = 60^{\circ}$.By

Vedclass · Accepted Answer

$50$

Vedclass · Answer

(D) Let $x$ be the height of the pole above the water surface.The angle of incidence $i$ with the normal is $90^{\circ} - 30^{\circ} = 60^{\circ}$.By Snell's law,$1 \cdot \sin 60^{\circ} = \mu_W \cdot \sin r$,where $r$ is the angle of refraction.$\sin r = \frac{\sin 60^{\circ}}{\mu_W} = \frac{\sqrt{3}/2}{4/3} = \frac{3\sqrt{3}}{8}$.Then,$\cos r = \sqrt{1 - \sin^2 r} = \sqrt{1 - \frac{27}{64}} = \sqrt{\frac{37}{64}} = \frac{\sqrt{37}}{8}$.$	an r = \frac{\sin r}{\cos r} = \frac{3\sqrt{3}/8}{\sqrt{37}/8} = \frac{3\sqrt{3}}{\sqrt{37}}$.From the geometry of the problem,the total shadow length $L$ is given by $L = x 	an i + h 	an r$,where $h = 1.5 \, m$ is the depth of water.Here,$	an i = 	an 60^{\circ} = \sqrt{3}$.So,$2.15 = x \sqrt{3} + 1.5 \cdot \frac{3\sqrt{3}}{\sqrt{37}}$.$x \sqrt{3} = 2.15 - \frac{4.5 \sqrt{3}}{\sqrt{37}} \approx 2.15 - \frac{4.5 	imes 1.732}{6.083} \approx 2.15 - 1.281 = 0.869$.$x = \frac{0.869}{\sqrt{3}} \approx \frac{0.869}{1.732} \approx 0.5016 \, m$.Thus,$x \approx 50 \, cm$.

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