A compound microscope consists of an objective lens of focal length 2.0cm and an eyepiece of focal length 6.25cm separated by a distance of 15 cm. How far from the objective should an object be placed in order to obtain the final image at (a) the least distance of distinct vision (25 cm), and (b) at infinity? What is the magnifying power of the microscope in each case?
Focal length of the objective lens f1=2cm Focal length of the eyepiece,f2=6.25cm Distance between the objective lens and the eyepiece,d=15cm Least distance of distinct vision,d′=25cm Image distance for the eyepiece,v2=−25cm bject distance for the eyepiece = u2 According to the lens formula f21=v21−u21 6.251=−251−u21 u2=−5cm Image distance for the objective lens v1=d+u2=15−5=10cm According to the lens formula 1=v11−u11 21=101−u11 u2=−2.5cm The magnifying power of a compound microscope is given by the relation m=∣u1∣v1(1+f2d′) m=2.510(1+6.2525)=20 b)The final image is formed at infinity Image distance for the eyepiece v2=∞ According to the lens formula f21=v21−u21 According to the lens formula 6.251=∞1−u21 u2=−6.25cm Image distance for the objective lens v1=d+u2=10−6.25=8.75 Object distance for the objective len =u2 According to the lens formula f11=v11−u11 21=8.751−u11 u2=−2.59cm The magnifying power of a compound microscope is given by the relation: ∣u1∣v1(∣u2∣d′)=13.51
