Given:
\(\mathrm{y}=\frac{10 \mathrm{~L}}{\pi} \sin \left(\frac{\pi \mathrm{x}}{5 \mathrm{~L}}\right) \).....(i)
According to the question co-ordinates of the point where a horizontal ray becomes vertical after reflection.
Differentiating equation (i) w.r.t. to \(x\). we get:
\(\frac{\mathrm{d} \mathrm{y}}{\mathrm{dx}}=\frac{\ 10 \mathrm{~L}}{\pi} \cos \left(\frac{\pi \mathrm{x}}{5 \mathrm{~L}}\right) \times \frac{ \pi}{ \mathrm{5L}} \times 2\)
\(=2 \cos \left(\frac{\pi \mathrm{x}}{5 \mathrm{~L}}\right) \)
Since the horizontal ray becomes vertical \(\frac{\mathrm{dy}}{\mathrm{dx}}=1\)
So,
\(2 \cos \left(\frac{\pi \mathrm{x}}{5 \mathrm{~L}}\right)=1 \)
\(\Rightarrow \cos \left(\frac{\pi \mathrm{x}}{5 \mathrm{~L}}\right)=\frac{1}{2} \)
\(\Rightarrow \cos\left(\frac{\pi \mathrm{x}}{5 \mathrm{~L}}\right)=\cos \frac{\pi}{3} \)
\(\Rightarrow \left(\frac{\pi \mathrm{x}}{5 \mathrm{~L}}\right)=\frac{\pi}{3} \)
\(\mathrm{x}=\frac{5 \mathrm{~L}}{3} \)
Put the value of \(x\) in equation (i),
\(\mathrm{y}=\frac{10 \mathrm{~L}}{\pi} \operatorname{sin}\left(\frac{\pi \times 5 \mathrm{~L}}{3 \times 5 \mathrm{~L}}\right) \)
\(y=\frac{5 \sqrt{3}L}{\pi}\)