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(D) Let the focal length of the convex lens be $f$. The magnification $m$ produced by a lens is given by $m = \frac{f}{f+u}$,where $u$ is the object d

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$15$

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(D) Let the focal length of the convex lens be $f$. The magnification $m$ produced by a lens is given by $m = \frac{f}{f+u}$,where $u$ is the object distance.For a convex lens,if the image size is the same,the magnification $m$ must be $m = 1$ (virtual image) or $m = -1$ (real image).Case $1$: For a real image,$m = -1$. Using the formula $m = \frac{f}{f+u}$,we have $-1 = \frac{f}{f+u_1}$,which implies $f+u_1 = -f$,so $u_1 = -2f$. Given $u_1 = -20 \ cm$,we get $2f = 20 \ cm$,so $f = 10 \ cm$. However,this would mean the object is at $2f$,forming a real image of the same size.Case $2$: If the image size is the same for two different positions,one must be a real image and one must be a virtual image. For a convex lens,the magnification $m = \frac{v}{u}$. Since the image size is the same,$|m_1| = |m_2|$.Let the object distances be $u_1 = -20 \ cm$ and $u_2 = -10 \ cm$. The magnification is $m = \frac{f}{f+u}$.For $u_1 = -20$,$m_1 = \frac{f}{f-20}$. For $u_2 = -10$,$m_2 = \frac{f}{f-10}$.Since the image sizes are equal,$|m_1| = |m_2|$. Thus,$\left| \frac{f}{f-20} \right| = \left| \frac{f}{f-10} \right|$.This implies $\frac{f}{20-f} = \frac{f}{f-10}$ (since one image is real and one is virtual,one magnification is negative). $20-f = f-10 \implies 2f = 30 \implies f = 15 \ cm$.

Images of same size are formed by a convex lens when an object is placed either at $20 \ cm$ or $10 \ cm$ distance from the lens. The focal length of the lens is (in $cm$)

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