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(B) The magnification $m$ for a lens is given by $m = \frac{f}{f+u}$,where $f$ is the focal length and $u$ is the object distance (using sign conventi

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$0.10$

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(B) The magnification $m$ for a lens is given by $m = \frac{f}{f+u}$,where $f$ is the focal length and $u$ is the object distance (using sign convention $u$ is negative,but here we use the formula $m = \frac{f}{f+u}$ where $u$ is the distance magnitude).Given $m_1 = 2m_2$ for $u_1 = 0.15 \ m$ and $u_2 = 0.2 \ m$.Substituting the values: $\frac{f}{f-0.15} = 2 	imes \frac{f}{f-0.2}$.Canceling $f$ from both sides: $\frac{1}{f-0.15} = \frac{2}{f-0.2}$.Cross-multiplying: $f - 0.2 = 2(f - 0.15)$.$f - 0.2 = 2f - 0.3$.$0.3 - 0.2 = 2f - f$.$f = 0.1 \ m$.

For a given lens,the magnification was found to be twice as large when the object was $0.15 \ m$ distant from it as when the distance was $0.2 \ m$. The focal length of the lens is.......$m$.

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