An object is placed at a distance of 1.50 , m | vedclass

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(C) Let the distance of the lens from the object be $x$. Then the distance of the lens from the screen is $(1.5 - x)$.Since the image is formed on the

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$0.3$

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(C) Let the distance of the lens from the object be $x$. Then the distance of the lens from the screen is $(1.5 - x)$.Since the image is formed on the screen,it is a real image. For a real image,the magnification $m = -4$.Using the magnification formula $m = \frac{v}{u}$,where $v = (1.5 - x)$ and $u = -x$:$-4 = \frac{1.5 - x}{-x}$$4x = 1.5 - x$$5x = 1.5$$x = 0.3 \, m$.Thus,the lens is placed at a distance of $0.3 \, m$ from the object.

An object is placed at a distance of $1.50 \, m$ from a screen. $A$ convex lens is placed between them to form a real image four times larger than the object on the screen. The position of the lens is at a distance of ..... (in $, m$)

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