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.

Answer:

            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, \[P=\frac{1}{{{f}_{2}}}=\frac{1}{0.2}=5D\]