question_answer

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:

A)  \[f=\frac{d}{{{m}_{1}}-{{m}_{2}}}\]

B)                        \[f=\frac{d}{{{m}_{1}}+{{m}_{2}}}\]        

C)        \[f=\frac{{{m}_{1}}-{{m}_{2}}}{d}\]         

D)         either (a)  and (b)

Correct Answer: A

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 \[\upsilon -u=d\]                 \[\therefore \]  \[f=\frac{d}{{{m}_{1}}-{{m}_{2}}}\]