A convex lens of focal length f produces a re | vedclass

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(D) Given that the magnification $m = -n$ (since the image is real).By definition of magnification,$m = \frac{v}{u} = -n$,which implies $v = -nu$.Usin

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$f(1+\frac{1}{n})$

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(D) Given that the magnification $m = -n$ (since the image is real).By definition of magnification,$m = \frac{v}{u} = -n$,which implies $v = -nu$.Using the lens formula: $\frac{1}{f} = \frac{1}{v} - \frac{1}{u}$.Substituting the values: $\frac{1}{f} = \frac{1}{-nu} - \frac{1}{u}$.$\frac{1}{f} = \frac{-1 - n}{nu} = -\frac{n+1}{nu}$.Since $f$ is positive for a convex lens,we consider the magnitude of the object distance $u$ (where $u$ is negative,so let $u = -x$):$\frac{1}{f} = \frac{n+1}{nx} \implies x = f(\frac{n+1}{n}) = f(1 + \frac{1}{n})$.Thus,the distance of the object from the lens is $f(1 + \frac{1}{n})$.

$A$ convex lens of focal length $f$ produces a real image whose size is $n$ times the size of an object. The distance of the object from the lens is

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