A point source of light S,placed at a distanc | vedclass

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(C) Let the mirror be placed along the $y$-axis with its centre at the origin $(0,0)$. The mirror extends from $y = -25 \,cm$ to $y = +25 \,cm$.The li

Vedclass · Accepted Answer

$150$

Vedclass · Answer

(C) Let the mirror be placed along the $y$-axis with its centre at the origin $(0,0)$. The mirror extends from $y = -25 \,cm$ to $y = +25 \,cm$.The light source $S$ is at $(60 \,cm, 0)$.The image $S'$ of the source $S$ in the plane mirror is formed at $(-60 \,cm, 0)$.The man walks along the line $x = 120 \,cm - 60 \,cm = 60 \,cm$ relative to the mirror plane,or more simply,the man is at a distance of $1.2 \,m = 120 \,cm$ from the mirror.The rays from the image $S'$ that define the field of view pass through the edges of the mirror at $y = 25 \,cm$ and $y = -25 \,cm$.Using similar triangles,the ratio of the height to the distance from the image $S'$ is constant:$\frac{y_{edge}}{distance_{S'}} = \frac{x}{distance_{S'} + distance_{man}}$Here,the distance of the image $S'$ from the mirror is $60 \,cm$. The distance of the man from the mirror is $120 \,cm$.So,the total distance of the man from the image $S'$ is $60 \,cm + 120 \,cm = 180 \,cm$.Using the property of similar triangles:$\frac{25 \,cm}{60 \,cm} = \frac{x}{180 \,cm}$$x = \frac{25 	imes 180}{60} = 25 	imes 3 = 75 \,cm$.This $x$ is the distance from the central axis to the extreme point on one side.The total distance between the two extreme points is $2x = 2 	imes 75 \,cm = 150 \,cm$.

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