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