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(C) The magnification $m$ for a lens is given by $m = \frac{f}{f+u}$.Since the numerical value of magnification remains the same,we have $|m_1| = |m_2

Vedclass · Accepted Answer

$35$

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(C) The magnification $m$ for a lens is given by $m = \frac{f}{f+u}$.Since the numerical value of magnification remains the same,we have $|m_1| = |m_2|$.For a real image,$m$ is negative,and for a virtual image,$m$ is positive. If the magnification remains the same numerically,we can write $\frac{f}{f+u_1} = -\frac{f}{f+u_2}$.Given $u_1 = -40 \,cm$ and $u_2 = -30 \,cm$.Substituting these values: $\frac{f}{f-40} = -\frac{f}{f-30}$.Since $f 
eq 0$,we divide by $f$: $\frac{1}{f-40} = -\frac{1}{f-30}$.Cross-multiplying gives: $f - 30 = -(f - 40)$.$f - 30 = -f + 40$.$2f = 70$.$f = 35 \,cm$.

If a lens is moved towards the object from a distance of $40 \,cm$ to $30 \,cm$,the magnification of the image remains the same (numerically). The focal length of the lens is ......... $cm$.

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