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(D) The incident ray is horizontal (angle $0^{\circ}$ with the $x$-axis) and the reflected ray is vertical (angle $90^{\circ}$ with the $x$-axis).Acco

Vedclass · Accepted Answer

Both $(B)$ and $(C)$

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(D) The incident ray is horizontal (angle $0^{\circ}$ with the $x$-axis) and the reflected ray is vertical (angle $90^{\circ}$ with the $x$-axis).According to the law of reflection,the angle of incidence equals the angle of reflection. If the normal makes an angle $	heta$ with the $x$-axis,the incident ray makes an angle $	heta$ with the normal and the reflected ray makes an angle $	heta$ with the normal.The total angle between the incident and reflected ray is $90^{\circ}$. Thus,the normal must make an angle of $45^{\circ}$ with the $x$-axis.The slope of the tangent to the surface is $\frac{dy}{dx} = 	an(	heta_{tangent})$. Since the normal makes $45^{\circ}$ with the $x$-axis,the tangent makes either $45^{\circ} + 90^{\circ} = 135^{\circ}$ or $45^{\circ} - 90^{\circ} = -45^{\circ}$ with the $x$-axis.$\frac{dy}{dx} = \frac{2L}{\pi} \cdot \frac{\pi}{L} \cos \left( \frac{\pi x}{L} ight) = 2 \cos \left( \frac{\pi x}{L} ight)$.Case $1$: $2 \cos \left( \frac{\pi x}{L} ight) = 	an(135^{\circ}) = -1 \implies \cos \left( \frac{\pi x}{L} ight) = -1/2 \implies \frac{\pi x}{L} = 120^{\circ} = \frac{2\pi}{3} \implies x = \frac{2L}{3}$.At $x = \frac{2L}{3}$,$y = \frac{2L}{\pi} \sin \left( \frac{2\pi}{3} ight) = \frac{2L}{\pi} \cdot \frac{\sqrt{3}}{2} = \frac{\sqrt{3} L}{\pi}$.Case $2$: $2 \cos \left( \frac{\pi x}{L} ight) = 	an(-45^{\circ}) = -1 \implies x = \frac{2L}{3}$ (same as above).Wait,if the slope is $1$,$\cos \left( \frac{\pi x}{L} ight) = 1/2 \implies \frac{\pi x}{L} = 60^{\circ} = \frac{\pi}{3} \implies x = \frac{L}{3}$.At $x = \frac{L}{3}$,$y = \frac{2L}{\pi} \sin \left( \frac{\pi}{3} ight) = \frac{\sqrt{3} L}{\pi}$.Both points $(L/3, \sqrt{3}L/\pi)$ and $(2L/3, \sqrt{3}L/\pi)$ satisfy the condition. Thus,the correct option is $(D)$.

$A$ reflecting surface is represented by the equation $y = \frac{2L}{\pi} \sin \left( \frac{\pi x}{L} \right)$,$0 \leq x \leq L$. $A$ ray travelling horizontally becomes vertical after reflection. The coordinates of the point$(s)$ where this ray is incident is?

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