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Question

A telescope has an objective of focal length 50 cm and eyepiece of focal length 5 cm. The least distance of distinct vision is 25 cm. The telescope is focussed for distinct vision on a scale 200 cm away from the objective. Calculate the magnitude of magnification produced.

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Solution

Given u=200cm,f=50cm
For image I1 of object formed by objective lens,
1f=1v1u
We have
1v=1f+1u=150+1200=41200=3200
v=+2003cm
Also, magnification produced by objective lens
m0=vu=200/3200=13
Image I1 acts as an object for eye lens.
Here, v=25cm,f=5cm
1f=1v1u
1u=1v=1f=12515=1+525
u=256cm
And magnification produced by eye lens,
me=vu=25(25/6)=6
a. The separation between objective and eyepiece
=|V|+|u|=2003+256=4256=70.73cm
b. Magnification produced, m=m0×me=13×6=2
The negative sign shown that the final image is inverted.

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