Two point sources S_1 and S_2 are 24 cm apar | vedclass

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(C) Let the distance of the lens from $S_1$ be $x$. Then the distance from $S_2$ is $(24 - x)$.For the images to form at the same place,one source mus

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$6$

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(C) Let the distance of the lens from $S_1$ be $x$. Then the distance from $S_2$ is $(24 - x)$.For the images to form at the same place,one source must form a virtual image and the other a real image.Let $S_1$ be at distance $u_1 = -x$ and $S_2$ be at distance $u_2 = +(24 - x)$.For $S_1$,the image $v$ is virtual,so $v = -v_0$. Using the lens formula $\frac{1}{v} - \frac{1}{u} = \frac{1}{f}$:$\frac{1}{-v_0} - \frac{1}{-x} = \frac{1}{9} \Rightarrow \frac{1}{v_0} = \frac{1}{x} - \frac{1}{9} \quad (i)$For $S_2$,the image $v$ is real,so $v = +v_0$. Using the lens formula:$\frac{1}{v_0} - \frac{1}{24-x} = \frac{1}{9} \Rightarrow \frac{1}{v_0} = \frac{1}{9} + \frac{1}{24-x} \quad (ii)$Equating $(i)$ and $(ii)$:$\frac{1}{x} - \frac{1}{9} = \frac{1}{9} + \frac{1}{24-x}$$\frac{1}{x} - \frac{1}{24-x} = \frac{2}{9}$$\frac{24-x-x}{x(24-x)} = \frac{2}{9} \Rightarrow \frac{24-2x}{24x-x^2} = \frac{2}{9}$$9(12-x) = 24x - x^2 \Rightarrow 108 - 9x = 24x - x^2$$x^2 - 33x + 108 = 0$$(x - 27)(x - 6) = 0$Since $x < 24$,we have $x = 6 \ cm$.

Two point sources $S_1$ and $S_2$ are $24 \ cm$ apart. Where should a convex lens of focal length $9 \ cm$ be placed in between them,so that the images of both sources are formed at the same place (in $cm$)?

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