Light incident normally on a plane mirror att | vedclass

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Question

(D) Given:Angle of deflection of the mirror,$	heta = 3.5^{\circ}$.Distance of the screen from the mirror,$D = 1.5\; m$.When a plane mirror is rotated

Vedclass · Accepted Answer

$18.4$

Vedclass · Answer

(D) Given:Angle of deflection of the mirror,$	heta = 3.5^{\circ}$.Distance of the screen from the mirror,$D = 1.5\; m$.When a plane mirror is rotated by an angle $	heta$,the reflected ray rotates by an angle $2	heta$.Therefore,the angle of deflection of the reflected ray is $2	heta = 2 	imes 3.5^{\circ} = 7.0^{\circ}$.The displacement $(d)$ of the reflected spot of light on the screen is given by the relation:$	an(2	heta) = \frac{d}{D}$Substituting the values:$d = D 	imes 	an(7.0^{\circ})$$d = 1.5\; m 	imes 	an(7.0^{\circ})$Using $	an(7.0^{\circ}) \approx 0.12278$$d = 1.5 	imes 0.12278 = 0.18417\; m$Converting the displacement into centimeters:$d = 0.18417 	imes 100\; cm = 18.417\; cm \approx 18.4\; cm$.Thus,the displacement of the reflected spot of light is $18.4\; cm$.

Light incident normally on a plane mirror attached to a galvanometer coil retraces its path as shown in the figure. $A$ current in the coil produces a deflection of $3.5^{\circ}$ of the mirror. What is the displacement (in $cm$) of the reflected spot of light on a screen placed $1.5\; m$ away?

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