A convergent lens is placed 40 ,cm to the rig | vedclass

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(C) For a parallel beam of light to emerge parallel from a system of two lenses, the second focal point of the first lens must coincide with the first

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$55$

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(C) For a parallel beam of light to emerge parallel from a system of two lenses, the second focal point of the first lens must coincide with the first focal point of the second lens.Let $L_1$ be the diverging lens with focal length $f_1 = -15 \,cm$ and $L_2$ be the convergent lens with focal length $f_2$.The parallel beam incident on $L_1$ appears to diverge from its focal point $F_1$, which is $15 \,cm$ to the left of $L_1$.For the beam to emerge parallel from $L_2$, the light rays incident on $L_2$ must appear to come from its focal point $F_2$, which is located to the left of $L_2$ at a distance equal to its focal length $f_2$.From the geometry of the system, the distance between the two lenses is $d = 40 \,cm$.The focal point $F_1$ is $15 \,cm$ to the left of $L_1$. Thus, the distance of $F_1$ from $L_2$ is $15 \,cm + 40 \,cm = 55 \,cm$.Since $F_1$ and $F_2$ must coincide for the rays to emerge parallel, the focal length of the convergent lens $f_2$ must be $55 \,cm$.

$A$ convergent lens is placed $40 \,cm$ to the right of a diverging lens of focal length $15 \,cm$. $A$ parallel beam of light enters the diverging lens from the left, and the beam is again parallel when it emerges from the convergent lens. The focal length of the convergent lens is (in $\,cm$)

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