A convex lens A of focal length 20 ,cm and a | vedclass

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(B) The focal length of the convex lens is $f_1 = 20 \,cm$ and the focal length of the concave lens is $f_2 = -56 \,cm$.For a combination of two thin

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$36$

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(B) The focal length of the convex lens is $f_1 = 20 \,cm$ and the focal length of the concave lens is $f_2 = -56 \,cm$.For a combination of two thin lenses separated by a distance $d$, the equivalent focal length $F$ is given by the formula:$\frac{1}{F} = \frac{1}{f_1} + \frac{1}{f_2} - \frac{d}{f_1 f_2}$Since the incident beam is parallel and the emergent beam is also parallel, the combination acts as a system with infinite focal length, i.e., $F = \infty$, which implies $\frac{1}{F} = 0$.Substituting the values into the formula:$0 = \frac{1}{20} + \frac{1}{-56} - \frac{d}{20 	imes (-56)}$$\frac{d}{20 	imes 56} = \frac{1}{20} - \frac{1}{56}$$\frac{d}{1120} = \frac{56 - 20}{1120}$$d = 56 - 20 = 36 \,cm$.Thus, the magnitude of the distance $d$ is $36 \,cm$.

$A$ convex lens $A$ of focal length $20 \,cm$ and a concave lens $B$ of focal length $56 \,cm$ are kept along the same axis with a distance $d$ between them. If a parallel beam of light falling on $A$ leaves $B$ as a parallel beam, then the magnitude of distance $d$ (in $cm$) is

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