• # question_answer A real image is formed by the lens at a distance of 20 cm from the lens. The image shifts towards the combination by 10 cm when a second lens is brought in contact with the first lens. Calculate the power of second lens.

Since, the image is real, so the lens is convex. Let its focal length be ${{f}_{1}}.$ It forms image on the other side So,                        $v=+20cm$             Using lens formula, $\frac{1}{v}-\frac{1}{u}=\frac{1}{f},$ we have                         $\frac{1}{20}-\frac{1}{u}=\frac{1}{{{f}_{1}}}$                       ?(i) When the second lens is placed in contact with the first lens, image shifts 10 cm closer to the combination, so the second lens is also convex. If ${{f}_{2}}$ is the focal length of second lens and f is the focal length of combination, then            $\frac{1}{f}=\frac{1}{{{f}_{1}}}+\frac{1}{{{f}_{2}}}$ For the combination, $v=20-10=+10\,cm$ So, from lens formula                         $\frac{1}{10}-\frac{1}{u}=\frac{1}{f}=\frac{1}{{{f}_{1}}}+\frac{1}{{{f}_{2}}}$        ?(ii) Subtracting Eq, (i) from Eq. (ii), we get                         $\frac{1}{{{f}_{2}}}=\frac{1}{10}-\frac{1}{20}=\frac{1}{20}$                         ${{f}_{2}}=20cm=0.2m$ $\therefore$  Power of second lens, $P=\frac{1}{{{f}_{2}}}=\frac{1}{0.2}=5D$
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