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(A) Let the distance of the object from the lens be $u$ and the distance of the image (screen) from the lens be $v$. Since the image is formed on a sc

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

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(A) Let the distance of the object from the lens be $u$ and the distance of the image (screen) from the lens be $v$. Since the image is formed on a screen, it is a real image.Given that the distance between the object and the screen is $x$, we have $v + u = x$.The magnification $m$ is given by $m = |v/u| = v/u$, so $v = mu$.Substituting $v = mu$ into the distance equation: $mu + u = x$, which gives $u(m + 1) = x$, or $u = x / (m + 1)$.Then, $v = x - u = x - x / (m + 1) = mx / (m + 1)$.Using the lens formula $\frac{1}{f} = \frac{1}{v} - \frac{1}{u}$:$\frac{1}{f} = \frac{1}{mx / (m + 1)} - \frac{1}{x / (m + 1)} = \frac{m + 1}{mx} - \frac{m + 1}{x} = \frac{m + 1 - m(m + 1)}{mx} = \frac{m + 1 - m^2 - m}{mx} = \frac{1 - m^2}{mx}$.Wait, for a real image, $v$ is positive and $u$ is negative. The lens formula is $\frac{1}{f} = \frac{1}{v} - \frac{1}{u}$. With $v = mu$ and $u = -u_0$ (where $u_0$ is distance), $m = v/u_0$. Thus $v = mu_0$ and $u_0 + v = x$. $u_0(1+m) = x \Rightarrow u_0 = x/(1+m)$ and $v = mx/(1+m)$.$\frac{1}{f} = \frac{1}{v} + \frac{1}{u_0} = \frac{1+m}{mx} + \frac{1+m}{x} = \frac{1+m + m(1+m)}{mx} = \frac{(1+m)^2}{mx}$.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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