The distance between an object and a divergen | vedclass

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(D) For a divergent (concave) lens, the focal length $f$ is taken as negative, so $f_{lens} = -f$ (where $f > 0$).Given the object distance $u = -mf$.

Vedclass · Accepted Answer

$\frac{1}{m+1}$

Vedclass · Answer

(D) For a divergent (concave) lens, the focal length $f$ is taken as negative, so $f_{lens} = -f$ (where $f > 0$).Given the object distance $u = -mf$.Using the lens formula: $\frac{1}{v} - \frac{1}{u} = \frac{1}{f_{lens}}$.Substituting the values: $\frac{1}{v} - \frac{1}{-mf} = \frac{1}{-f}$.$\frac{1}{v} + \frac{1}{mf} = -\frac{1}{f}$.$\frac{1}{v} = -\frac{1}{f} - \frac{1}{mf} = -\frac{1}{f} \left(1 + \frac{1}{m}\right) = -\frac{1}{f} \left(\frac{m+1}{m}\right)$.Therefore, $v = -f \left(\frac{m}{m+1}\right)$.The linear magnification $M$ is given by $M = \frac{v}{u}$.$M = \frac{-f \left(\frac{m}{m+1}\right)}{-mf} = \frac{1}{m+1}$.

The distance between an object and a divergent lens is $m$ times the focal length of the lens. The linear magnification produced by the lens is

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