When the object is at distances u_1 and u_2 f | vedclass

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(B) Let the focal length of the lens be $f$. The magnification $m$ is given by $m = \frac{v}{u}$.For a real image at distance $u_1$,the magnification

Vedclass · Accepted Answer

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

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(B) Let the focal length of the lens be $f$. The magnification $m$ is given by $m = \frac{v}{u}$.For a real image at distance $u_1$,the magnification is $m_1 = \frac{f}{f - u_1}$. Since the image is real,$m_1$ is negative,so $m_1 = -\frac{f}{u_1 - f}$.For a virtual image at distance $u_2$,the magnification is $m_2 = \frac{f}{f - u_2}$. Since the image is virtual,$m_2$ is positive,so $m_2 = \frac{f}{f - u_2}$.Given that the sizes are the same,$|m_1| = |m_2|$.Thus,$|-\frac{f}{u_1 - f}| = |\frac{f}{f - u_2}|$.Since $u_1 > f$ (for real image) and $u_2 This simplifies to $f - u_2 = u_1 - f$.Rearranging the terms,we get $2f = u_1 + u_2$,which implies $f = \frac{u_1 + u_2}{2}$.

When the object is at distances $u_1$ and $u_2$ from a lens,the images formed are real and virtual respectively,and both have the same size. Then the focal length of the lens is:

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