Q: Find the magnifying power of a compound microscope whose objective has a focal power of 100 D and eye piece has a focal power of 16 D when the object is placed at a distance of 1.1 cm from the objective. Assume that the final image is formed at the least distance of distinct vision (25 cm)

Sol: The magnifying power of a compound microscope when the final image forms at the least distance of

distinct vision,

$\large m = \frac{v_o}{u_o} (1 + \frac{D}{f_e})$

u_{o} = -1.1 cm , $f_o = \frac{1}{P_o}m = \frac{100}{P_o} cm$

$f_o = \frac{100}{P_o} cm = \frac{100}{100} = 1 cm$

$\large \frac{1}{v_o} – \frac{1}{u_o} = \frac{1}{f_o} $

$\large \frac{1}{v_o} – \frac{1}{-1.1} = \frac{1}{1} $

v_{o} = 11 cm

$f_e = \frac{1}{P_e}m = \frac{100}{P_e} cm$

$f_e = \frac{100}{P_o} cm = \frac{100}{16} = 6.25 cm$

Least distance of distinct vision, D = 25 cm

$\large m = \frac{11}{-1.1} (1 + \frac{25}{6.25})$

m = – 50