A convex lens produces a real image m times t | vedclass

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(A) For a convex lens, the magnification $m$ is given by $m = \frac{v}{u}$.Since the image is real, the magnification is negative, so $m = -|m|$.The l

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$\left( \frac{m + 1}{m} \right)f$

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(A) For a convex lens, the magnification $m$ is given by $m = \frac{v}{u}$.Since the image is real, the magnification is negative, so $m = -|m|$.The lens formula is $\frac{1}{f} = \frac{1}{v} - \frac{1}{u}$.Substituting $v = mu$ into the lens formula: $\frac{1}{f} = \frac{1}{mu} - \frac{1}{u} = \frac{1 - m}{mu}$.Rearranging for $u$: $u = \frac{f(1 - m)}{m} = -\left( \frac{m - 1}{m} \right)f$.Since the distance is the magnitude of $u$, the distance is $\left( \frac{m - 1}{m} \right)f$ if $m$ is the magnification magnitude. However, using the standard sign convention where $m$ is the ratio of image size to object size, for a real image, $v = -mu$ (where $m$ is positive magnitude). Substituting into $\frac{1}{f} = \frac{1}{v} - \frac{1}{u}$ gives $\frac{1}{f} = \frac{1}{-mu} - \frac{1}{u} = -\frac{1+m}{mu}$.Thus, $u = -\left( \frac{m+1}{m} \right)f$. The distance is $\left( \frac{m+1}{m} \right)f$.

$A$ convex lens produces a real image $m$ times the size of the object. What will be the distance of the object from the lens?

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