Question

Find the maximum magnifying power of a microscope having a 25D lens as the objective, a 5 D lens as the eye-piece and the separation 30cm between the two lenses. The least distance for clear vision is 25cm.

Answer 1

Maximum magnifying power will be in the case when final image formed at the least distance of distinct vision.

M = \(-\frac{v_o}{u_o}(1+\frac{D}{f_e})\)

Separation between the lenses in this case |v_o | + |u_o | = 30cm

v_e = D = 25cm

Use \(\frac{1}{f_e}=\frac{1}{v_e}-\frac{1}{u_e}\) here \(f_e=\frac{1}{p}=\frac{1}{5}m\) = 20cm

\(\frac{1}{20}=\frac{1}{-25}-\frac{1}{u_e}\)

or \(\frac{-1}{u_e}=\frac{1}{20}+\frac{1}{25}=\frac{5+4}{100}\)

or u_e = -100/9cm

but |v_o | + |u_e | = 30cm

So |v_o | = 30 − |u_e | = \(30-\frac{100}{9}=\frac{170}{9}cm\)

Now, for objective

\(\frac{1}{f_o}=\frac{1}{v_o}-\frac{1}{u_o}\) here \(f_o=\frac{1}{p}=\frac{1}{25}m\) = 4 cm

\(\frac{1}{4}=\frac{1}{170/9}-\frac{1}{u_e}\)

\(\frac{1}{u_o}=-\frac{9}{170}-\frac{1}{4}=\frac{360-170}{4\times170}\)

or \(u_o=-\frac{4\times 170}{134}=\frac{340}{67}cm\)

So, magnifying power

M = \(-\frac{\frac{170}{9}}{\frac{340}{67}}(1+\frac{25}{20})=-8.4.\)