Question

a person with normal near point 25cm using a compound microscope with objective of focal length 8 mm and an eyepiece of focal length 2.5 CM can bring an object at 9 mm from the objective in same focus separation between the two lenses calculate the magnifying power​

Answer 1

Answer:

Given :

d=25cm

f0=8.0mm=0.8cm

fe=2.5cm

u0=−9.0mm

=−0.9cm

1ve−1ue=1fe

=> 1ue=1ve−1fe

=1−25−12.5

=−1−1025

=1125

ue=−2511

=−2.27cm

Also, 1v0−1u0=1f0

1v0=1f0+1v0

=10.8+1−0.9

=0.9−0.80.72

=7.2cm

Seperation between two lenses =|ue|+v0=2.27+7.2=9.47cm

hope it will help

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