A diverging lens with a focal length magnitud | vedclass

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

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

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(B) For the diverging lens,the incident light is parallel,so the image is formed at its focus. Since it is a diverging lens,$f_1 = -25\ cm$.Using the lens formula $\frac{1}{v_1} - \frac{1}{u_1} = \frac{1}{f_1}$,with $u_1 = \infty$,we get $v_1 = f_1 = -25\ cm$.This image acts as a virtual object for the converging lens. The distance between the lenses is $d = 15\ cm$.The object distance for the converging lens is $u_2 = -(25 + 15) = -40\ cm$.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 magnitude of $25\ cm$ is placed at a distance of $15\ cm$ from a converging lens with a focal length magnitude of $20\ cm$. $A$ beam of parallel light falls on the diverging lens. The final image formed is:

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