Q. If $ {{m}_{1}} $ and $ {{m}_{2}} $ be the linear magnification of the object in the conjugate positions of a convex lens, and if d be the distance between the conjugate positions, then the focal length of the lens is given by:
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Solution:
For the first position $ {{m}_{1}}=\frac{\upsilon }{u} $
For the second position $ {{m}_{2}}=\frac{u}{\upsilon } $ From the lens formula $ f=\frac{u\upsilon }{u+\upsilon } $ or $ f=\frac{u\upsilon }{u+\upsilon }\times \frac{\upsilon -u}{\upsilon -u} $ $ =\frac{(\upsilon -u)}{({{\upsilon }^{2}}-{{u}^{2}})/u\upsilon } $ i.e., $ f=\frac{u-\upsilon }{\left( \frac{\upsilon }{u}-\frac{u}{\upsilon } \right)} $ But $ \upsilon -u=d $ $ \therefore $ $ f=\frac{d}{{{m}_{1}}-{{m}_{2}}} $
