The minimum distance between an object and it | vedclass

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Question

(D) Let the distance of the object from the convex lens be $u$ and the distance of the real image be $v$. Both $u$ and $v$ are positive in magnitude.

Vedclass · Accepted Answer

$4$

Vedclass · Answer

(D) Let the distance of the object from the convex lens be $u$ and the distance of the real image be $v$. Both $u$ and $v$ are positive in magnitude. The total distance $D$ between the object and the image is $D = u + v$.From the lens formula,$\frac{1}{f} = \frac{1}{v} - \frac{1}{u}$. Using sign convention,$u$ is negative,so $\frac{1}{f} = \frac{1}{v} - \frac{1}{-u} = \frac{1}{v} + \frac{1}{u}$.Thus,$v = \frac{uf}{u-f}$.The total distance is $D = u + \frac{uf}{u-f} = \frac{u^2 - uf + uf}{u-f} = \frac{u^2}{u-f}$.To find the minimum distance,we differentiate $D$ with respect to $u$ and set it to zero: $\frac{dD}{du} = \frac{(u-f)(2u) - u^2(1)}{(u-f)^2} = 0$.$2u^2 - 2uf - u^2 = 0 \implies u^2 - 2uf = 0 \implies u = 2f$.Substituting $u = 2f$ into the expression for $v$,we get $v = \frac{2f \cdot f}{2f - f} = 2f$.Therefore,the minimum distance $D = u + v = 2f + 2f = 4f$.

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

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