The minimum distance between an object and it | vedclass

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Question

(D) Let the object distance be $u$ (where $u  0$). The lens formula is given by:$\frac{1}{v} - \frac{1}

Vedclass · Accepted Answer

$4$

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(D) Let the object distance be $u$ (where $u  0$). The lens formula is given by:$\frac{1}{v} - \frac{1}{u} = \frac{1}{f}$Let the distance between the object and the real image be $L$. Since the object is on the left $(u  0)$,the distance $L$ is given by:$L = v + |u| = v - u$From the lens formula,$\frac{1}{v} = \frac{1}{f} + \frac{1}{u} = \frac{u+f}{uf}$,so $v = \frac{uf}{u+f}$.Substituting this into the expression for $L$:$L = \frac{uf}{u+f} - u = \frac{uf - u(u+f)}{u+f} = \frac{-u^2}{u+f}$Since $u$ is negative,let $u = -x$ where $x > 0$. Then $L = \frac{-(-x)^2}{-x+f} = \frac{x^2}{x-f}$.To find the minimum distance,we differentiate $L$ with respect to $x$ and set it to zero:$\frac{dL}{dx} = \frac{(x-f)(2x) - x^2(1)}{(x-f)^2} = 0$$2x^2 - 2xf - x^2 = 0 \Rightarrow x^2 - 2xf = 0$Since $x 
eq 0$,we have $x = 2f$.Substituting $x = 2f$ into the expression for $L$:$L_{	ext{min}} = \frac{(2f)^2}{2f-f} = \frac{4f^2}{f} = 4f$.

The minimum distance between an object and its real image formed by a convex lens of focal length $f$ is (in $f$)

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