The magnifications produced by a convex lens | vedclass

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(D) Given, $m_1 = 4$ and $m_2 = 3$.The distance between the two object positions is $|u_2 - u_1| = 2 \,cm$.For a real image formed by a convex lens, t

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$24$

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(D) Given, $m_1 = 4$ and $m_2 = 3$.The distance between the two object positions is $|u_2 - u_1| = 2 \,cm$.For a real image formed by a convex lens, the magnification is $m = \frac{f}{f+u}$. Since the image is real, $u$ is negative, so let $u = -x$. Then $m = \frac{f}{f-x}$.Rearranging gives $f-x = \frac{f}{m}$, or $x = f(1 - \frac{1}{m}) = f(\frac{m-1}{m})$.For $m_1 = 4$, $u_1 = f(\frac{4-1}{4}) = \frac{3f}{4}$.For $m_2 = 3$, $u_2 = f(\frac{3-1}{3}) = \frac{2f}{3}$.The separation is $u_1 - u_2 = 2 \,cm$ (assuming $m_1 > m_2$ implies $u_1 $\frac{3f}{4} - \frac{2f}{3} = 2$.$\frac{9f - 8f}{12} = 2$.$\frac{f}{12} = 2 \Rightarrow f = 24 \,cm$.

The magnifications produced by a convex lens for two positions of an object are $4$ and $3$, respectively. If the distance of separation between the two positions of the object is $2 \,cm$, then the focal length of the lens is (in $\,cm$)

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