A point source of light is placed at 2f from | vedclass

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Question

(C) Let $S$ be the point source placed at a distance $u = -2f$ from the converging lens.Using the lens formula $\frac{1}{v} - \frac{1}{u} = \frac{1}{f

Vedclass · Accepted Answer

$45$

Vedclass · Answer

(C) Let $S$ be the point source placed at a distance $u = -2f$ from the converging lens.Using the lens formula $\frac{1}{v} - \frac{1}{u} = \frac{1}{f}$,we have $\frac{1}{v} - \frac{1}{-2f} = \frac{1}{f}$,which gives $\frac{1}{v} = \frac{1}{f} - \frac{1}{2f} = \frac{1}{2f}$. Thus,$v = 2f$.This means the lens forms an image $S'$ at a distance $2f$ from the lens on the other side.For the rays reflected from the mirror to be parallel after passing through the lens again,the rays incident on the lens from the mirror must appear to come from the focus of the lens.Let the mirror be placed at distance $d$ from the lens. The rays from the lens converge towards $S'$ at distance $2f$. The mirror reflects these rays,creating a virtual object $S''$ for the lens. For the final rays to be parallel,the object $S''$ must be at the focus $F$ of the lens.From the geometry,the distance of the mirror from the lens is $d$. The distance of $S'$ from the lens is $2f$. The distance of $S''$ from the lens is $f$.The mirror is placed at the midpoint between $S''$ and $S'$. Therefore,$d = \frac{f + 2f}{2} = \frac{3f}{2}$.Given $f = 30 \, cm$,we have $d = \frac{3 	imes 30}{2} = 45 \, cm$.

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