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(B) Case $1$: When the object is at infinity,the image is formed at the focal point $F_2$. So,$v_1 = f$.Case $2$: The object is placed at $20 \ cm$ fr

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$10$

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(B) Case $1$: When the object is at infinity,the image is formed at the focal point $F_2$. So,$v_1 = f$.Case $2$: The object is placed at $20 \ cm$ from the object-side focal plane. This means the object distance $u = -(f + 20) \ cm$.Using the lens formula $\frac{1}{f} = \frac{1}{v} - \frac{1}{u}$:$\frac{1}{f} = \frac{1}{v_2} - \frac{1}{-(f + 20)} \Rightarrow \frac{1}{v_2} = \frac{1}{f} - \frac{1}{f + 20} = \frac{f + 20 - f}{f(f + 20)} = \frac{20}{f(f + 20)}$.Thus,$v_2 = \frac{f(f + 20)}{20}$.The distance between the two images is $|v_2 - v_1| = 5 \ cm$.$\frac{f^2 + 20f}{20} - f = 5 \Rightarrow f^2 + 20f - 20f = 100 \Rightarrow f^2 = 100 \Rightarrow f = 10 \ cm$.Alternatively,using Newton's form of the lens equation $x_1 x_2 = f^2$,where $x_1 = 20 \ cm$ and $x_2 = 5 \ cm$,we get $f^2 = 20 	imes 5 = 100$,so $f = 10 \ cm$.

An object is placed first at infinity and then at $20 \ cm$ from the object-side focal plane of a convex lens. The two images thus formed are $5 \ cm$ apart. The focal length of the lens is $...... \ cm$.

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