When an object is at a distance u_1 and u_2 f | vedclass

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Question

(C) The magnification $m$ for a lens is given by $m = \frac{f}{f + u}$.For a real image,the magnification is negative,so $m_1 = -m = \frac{f}{f - u_1}

Vedclass · Accepted Answer

$\frac{u_1 + u_2}{2}$

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(C) The magnification $m$ for a lens is given by $m = \frac{f}{f + u}$.For a real image,the magnification is negative,so $m_1 = -m = \frac{f}{f - u_1}$.For a virtual image,the magnification is positive,so $m_2 = +m = \frac{f}{f - u_2}$.Since the magnitudes of the magnifications are equal,we have $\frac{f}{f - u_1} = -\frac{f}{f - u_2}$.Canceling $f$ from both sides,we get $\frac{1}{f - u_1} = -\frac{1}{f - u_2}$.Cross-multiplying gives $f - u_2 = -(f - u_1) = -f + u_1$.Rearranging the terms,$2f = u_1 + u_2$,which leads to $f = \frac{u_1 + u_2}{2}$.

When an object is at a distance $u_1$ and $u_2$ from a lens,a real image and a virtual image are formed,respectively,having the same magnification $m$. The focal length of the lens is:

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