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Q.

If we need a magnification of 375 from a compound microscope of tube length 150mm and an objective of focal length 5mm, the focal length of the eye-piece, when the final image is at near point is  

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a

22mm

b

12mm

c

2mm

d

33mm

answer is A.

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Detailed Solution

According question,  M=375
L=150mm,f0=5mm  and  fe=? 
Using, magnification,  M≃Lf0(1+Dfe)
⇒375=1505(1+250fe)(∵D=25cm=250mm)     ⇒12.5=1+250fe⇒fe=25011.5=21.7≈22mm

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