CGPET 2014 Solved Paper

As, the final image in formed at infinity by the eye lens, so the object distance for eye lens is focal length of eye lens.
So, ‌‌ue=6‌mm=0.6‌cm
The distance between the objective and eye lens is 20.6‌cm.
∴‌‌L=v0+ue‌⇒v0=L−ue
‌=20.6−0.6
‌=20‌cm
Let u be the object distance form objective, then from lens formula for objective.
‌

1

f0

=‌

1

v0

−‌

1

u0

‌⇒‌

1

u0

=−‌

1

f0

+‌

1

v0

=‌

1

20

−‌

1

0.6

‌=−‌

19.4

12

‌ or ‌‌‌u0‌=−‌

12

19.4

=−0.62‌cm.
So, the magnification of objective,
m‌=‌

v0

u0

=‌

20

−0.62

‌=−32.3