The reflecting surface is given by $y = \frac{10 L}{π} sin \frac{π x}{5 L}$ . The co-ordinates of the point where a horizontal ray becomes vertical after reflection is:

\frac{5 L}{3}, \frac{5 \sqrt{3} L}{π}

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\frac{5 \sqrt{3} L}{π}, \frac{5 L}{3}

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\frac{5 L}{\sqrt{3}}, \frac{π L}{5}

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\frac{π L}{5}, \frac{5 L}{\sqrt{3}}

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Solution

The correct option is B $\frac{5 L}{3}, \frac{5 \sqrt{3} L}{π}$
$y = \frac{10 L}{π} s i n (\frac{π x}{5 L})$
$\frac{d y}{d x} = \frac{/ 10 L}{/ π} c o s (\frac{π x}{5 L}) \times \frac{/ π}{/ 5 L} \times 2$
Since the horizontal ray becomes vertical $\frac{d y}{d x} = 1$
$2 c o s (\frac{π x}{5 L}) = 1$
$(\frac{π x}{5 L}) = \frac{x}{3}$
$x = \frac{5 L}{3}$
$y = \frac{10 L}{π} s i n e (\frac{π \times / 5 / L}{3 \times / 5 / L})$
$= \frac{5 \sqrt{3 L}}{π}$

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