A convex lens of focal length f is placed som | vedclass

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(D) Let the distance of the object from the lens be $u$ and the distance of the image from the lens be $v$. Since the object and screen are fixed at a

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$\frac{m x}{(m+1)^2}$

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(D) Let the distance of the object from the lens be $u$ and the distance of the image from the lens be $v$. Since the object and screen are fixed at a distance $x$, we have $v + u = x$ (taking magnitudes).The magnification is given by $m = \frac{v}{u}$, so $v = mu$.Substituting $v = mu$ into the distance equation: $mu + u = x$, which gives $u(m + 1) = x$, so $u = \frac{x}{m+1}$.Then $v = x - u = x - \frac{x}{m+1} = \frac{mx}{m+1}$.Using the lens formula $\frac{1}{f} = \frac{1}{v} - \frac{1}{u}$, where $u$ is negative by sign convention $(u = -\frac{x}{m+1})$:$\frac{1}{f} = \frac{1}{(\frac{mx}{m+1})} - \frac{1}{(-\frac{x}{m+1})}$$\frac{1}{f} = \frac{m+1}{mx} + \frac{m+1}{x} = \frac{m+1}{x} (\frac{1}{m} + 1) = \frac{m+1}{x} (\frac{1+m}{m}) = \frac{(m+1)^2}{mx}$.Therefore, $f = \frac{mx}{(m+1)^2}$.

$A$ convex lens of focal length $f$ is placed somewhere in between an object and a screen. The distance between the object and the screen is $x$. If the numerical value of the magnification produced by the lens is $m$, then the focal length of the lens is

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