wiz-icon
MyQuestionIcon
MyQuestionIcon
1
You visited us 1 times! Enjoying our articles? Unlock Full Access!
Question

The length of a compound microscope is 14 cm and its magnifying power when final image is formed at near point is 25. If the focal length of eyepiece is 5 cm, the focal length of objective lens is:

A
5931 cm
No worries! We‘ve got your back. Try BYJU‘S free classes today!
B
5925 cm
No worries! We‘ve got your back. Try BYJU‘S free classes today!
C
2 cm
No worries! We‘ve got your back. Try BYJU‘S free classes today!
D
1.5 cm
Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses
Open in App
Solution

The correct option is D 1.5 cm
L=14 cm,m=25Fe=5 cm we know that L=v0+Fev0=LFe=9 cm
m=V0U0DFe[D=25 cm]25=940255u0=95 cm
and for objective len8:
1f0=1u0+1v0=59+19
F0=96=1.5 cm

flag
Suggest Corrections
thumbs-up
0
Join BYJU'S Learning Program
similar_icon
Related Videos
thumbnail
lock
Combination of Lenses and Mirrors
PHYSICS
Watch in App
Join BYJU'S Learning Program
CrossIcon