A compound microscope is used to enlarge an object kept at a distance 0.03m from it’s objective which consists of several convex lenses in contact and has focal length 0.02m. If a lens of focal length 0.1m is removed from the objective, then by what distance the eye-piece of the microscope must be moved to refocus the image (in cm)

Moderate

Solution

$\begin{array}{l} I f i n i t i a l l y t h e o b j e c t i v e (f o c a l l e n g t h f_{o}) f o r m s t h e i m a g e a t d i s \tan c e v_{o} t h e n \\ v_{o} = \frac{u_{o} f_{o}}{u_{o} - f_{o}} = \frac{3 \times 2}{3 - 2} = 6 c m \\ N o w a s i n c a s e o f l e n s e s i n c o n t a c t \\ \frac{1}{F_{o}} = \frac{1}{f_{1}} + \frac{1}{f_{2}} + \frac{1}{f_{3}} + ..... = \frac{1}{f_{1}} + \frac{1}{{F^{'}}_{o}} {w h e r e \frac{1}{{F^{'}}_{o}} = \frac{1}{f_{2}} + \frac{1}{f_{3}} + ....} \\ S o, i f o n e o f t h e l e n s i s r e m o v e d, t h e f o c a l l e n g t h o f t h e r e m a i n i n g l e n s s y s t e m \\ \frac{1}{{F^{'}}_{o}} = \frac{1}{F_{0}} - \frac{1}{f_{1}} = \frac{1}{2} - \frac{1}{10} \\ \Rightarrow {F^{'}}_{o} = 2.5 c m \\ T h i s l e n s w i l l f o r m t h e i m a g e o f s a m e o b j e c t a t a d i s \tan c e {v^{'}}_{o} s u c h t h a t \\ {v^{'}}_{o} = \frac{u_{o} {F^{'}}_{o}}{u_{o} - {F^{'}}_{o}} = \frac{3 \times 2.5}{(3 - 2.5)} = 15 c m \\ S o t o r e f o c u s t h e i m a g e, e y e - p i e c e m u s t b e m o v e d b y t h e s a m e d i s \tan c e t h r o u g h \\ w h i c h t h e i m a g e f o r m e d b y t h e o b j e c t i v e h a s s h i f t e d i . e . 15 - 6 = 9 c m \end{array}$

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