Question 91 of 295

A pole is vertically submerged in swimming pool, such that it gives a length of shadow $2.15m$ within water when sunlight is incident at an angle of $30^{\circ}$ with the surface of water. If swimming pool is filled to a height of $1.5m,$ then the height of the pole above the water surface in centimeters is ({n}_{w}=\dfrac{4}{3})________.

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Answer: 50

Let angle $BAC$ is $\theta$.
Applying Snell's law at $E$, we get $1\times \sin 60^{\circ}=\dfrac{4}{3}\times \sin \theta$
$\Rightarrow \sin \theta =\dfrac{3\sqrt{3}}{8} \Rightarrow \tan \theta =\dfrac{\sin \theta }{\sqrt{1-{\left(\sin \theta \right)}^{2}}}=\dfrac{3\sqrt{3}}{\sqrt{37}}$
Now, $BC=1.5\times \tan \theta =1.28$
Then, $AB=2.15-BC=0.86m$
Required value, $DF=AB\tan 30^{\circ}=0.5015m\approx 50cm$