At what distance from a biconvex lens of foca | vedclass

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(A) Let the distance of the object from the lens be $u$ and the distance of the real image from the lens be $v$. Since the image is real, $u$ and $v$

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$2 F$

Vedclass · Answer

(A) Let the distance of the object from the lens be $u$ and the distance of the real image from the lens be $v$. Since the image is real, $u$ and $v$ are positive in magnitude. The total distance $D$ between the object and the image is $D = u + v$.Using the lens formula, $\frac{1}{f} = \frac{1}{v} - \frac{1}{-u} = \frac{1}{v} + \frac{1}{u}$.Thus, $v = \frac{uf}{u-f}$.The total distance is $D = u + \frac{uf}{u-f}$.To find the minimum distance, we differentiate $D$ with respect to $u$ and set it to zero: $\frac{dD}{du} = 1 + \frac{f(u-f) - uf}{(u-f)^2} = 1 - \frac{f^2}{(u-f)^2} = 0$.This gives $(u-f)^2 = f^2$, so $u-f = f$ (since $u > f$), which means $u = 2F$.At $u = 2F$, the distance $D = 2F + 2F = 4F$, which is the minimum distance.

At what distance from a biconvex lens of focal length $F$ must an object be placed so that the distance between the object and its real image is minimal?

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