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

Answer

D

Explanation

For the given compound microscope, `f_0 = (1)/(25 diopter)` = 0.04 m = 4 cm, `f_0 = (1)/(5 diopter) = 0.2 m =20 cm` D = 25 cm, separation between objective and eyepiece = 30 cm. The magnifying power is maximum when the image if formed by the eye piece at the least distance of clear vision i.e, D =25 cm. For the eye piece, `V_e= -25 cm, f_e = 20 cm From lens formula
`(1)/(upsilon_e = (1)/(u_e) + (1)/(f_e)`
`rArr = (1)/(u_e) = (1)/(upsilon_e) - (1)/(f_e)`
` =(1)/(-25) - (1)/(20) = -((4 +5))/(100)`
`=u_e = (-100)/(9) = 11.11 cm`
So, the maximum magnifying power is given by,
` m = (upsilon_0)/(u_0) = (1 + (D)/(f_e))`
`=-((18.89)/(-5.07)) (1+ ((25)/(20)))`
3.7225xx 2.25 = 8.376