Question

A plane mirror hinged at O is free to rotate in a vertical plane. The point O is at a distance x from a long screen placed in front of the mirror as shown in figure. A laser beam of light incident vertically downward is reflected by he mirror at O so that a bright spot is formed at the screen.
At the instant shown, the angle of inclidence is $\theta$ and the mirror is rotating clockwise with constant angular velocity w. Find the speed of the spot at this instant.

• A. $2xwse{c}^2\varphi$
• B. $2xwco{s}^2\varphi$
• C. $2xwsi{n}^2\varphi$
• D. $2xwta{n}^2\varphi$

Answer 1

The correct option is A $2xwse{c}^2\varphi$


Let P be the bright spot, shown on the screen, Let the distance of point P from O' be y at this instant shown in figure
$\frac{dy}{dt}=?$
From the figure,



$\theta +\theta +\varphi ={90}^{\circ }\to \left(1\right)\theta +\varphi +\alpha ={90}^{\circ }\left(2\right)\Rightarrow \theta =\alpha \Rightarrow \varphi +2\theta ={90}^{\circ }\Rightarrow ‘\varphi +2\alpha ={90}^{\circ }\Rightarrow \frac{d\varphi }{dt}+2\frac{d\alpha }{dt}=0\Rightarrow \frac{d\varphi }{dt}=\frac{2d\alpha }{dt}$
So, the angular spped of the reflected ray is double the angular speed of the mirror since,
$y=xtan\varphi \Rightarrow \frac{dy}{dt}=sse{c}^2\varphi \frac{d\varphi }{dt}=2xwse{c}^2\varphi$