Draw a labelled ray diagram of a compound microscope and write an expression for its magnifying power. The focal lengths of the objective and eye-lens of a compound microscope are 2cm and 6.25cm respectively. The distance between the lenses is 15cm. (i) How far from the objective lens will the object be kept so as to obtain the final image at the near point of the eye? (ii) Also calculate its magnifying power.
Answer: (i) Object distance =2.5cm from the objective. (ii) Magnifying power =20.
Solution
Magnifying power (image at near point): m=foL(1+feD). \textbf{Eyepiece:} final image at ve=−25cm, fe=6.25cm: ue1=ve1−fe1=−251−6.251=−51⇒ue=−5cm. So the objective's image is 5cm before the eyepiece, i.e. vo=15−5=10cm. \textbf{Objective:} fo=2cm, vo=10cm: uo1=vo1−fo1=101−21=−52⇒uo=−2.5cm. (i) uo=2.5cm. (ii) m=∣uo∣vo(1+feD)=2.510(1+6.2525)=4×5=20.
