A diverging lens with a focal length of magni | vedclass

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(C) For the diverging lens,the incident light is parallel,so the image is formed at its focus. Since it is a diverging lens,$f_1 = -25 \, cm$. The ima

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real and at a distance of $40 \, cm$ from the converging lens

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(C) For the diverging lens,the incident light is parallel,so the image is formed at its focus. Since it is a diverging lens,$f_1 = -25 \, cm$. The image is virtual and formed at $v_1 = -25 \, cm$ from the diverging lens.This image acts as an object for the converging lens. The distance between the lenses is $d = 15 \, cm$. The object distance for the converging lens is $u_2 = -(25 + 15) \, cm = -40 \, cm$.For the converging lens,$f_2 = +20 \, cm$. Using the lens formula $\frac{1}{v_2} - \frac{1}{u_2} = \frac{1}{f_2}$:$\frac{1}{v_2} - \frac{1}{-40} = \frac{1}{20}$$\frac{1}{v_2} = \frac{1}{20} - \frac{1}{40} = \frac{2-1}{40} = \frac{1}{40}$$v_2 = +40 \, cm$.Since $v_2$ is positive,the final image is real and formed at a distance of $40 \, cm$ from the converging lens.

$A$ diverging lens with a focal length of magnitude $25 \, cm$ is placed at a distance of $15 \, cm$ from a converging lens with a focal length of magnitude $20 \, cm$. $A$ beam of parallel light falls on the diverging lens. The final image formed is:

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