(a) The separation between objective lens and the eyepiece can be calculated in both the conditions of most relaxed eye and most strained eye.
Most relaxed eye
L = fo+fe = 140 + 5 = 145 cm
Most strained eye object distance ‘ue’ for eye lens

1

ve

−

1

ue

=

1

fe

or

1

−25

−

1

ue

=

1

5

-

1

ue

=

1

5

+

1

25

=

6

25

or ue = -

25

6

cm = - 4.16 cm
Separation between lenses
L = fo+|ue| = 145 + 4.16 = 149.16 cm
(b) We can calculate height of image by objective lens

tan θ =

A′B′

B′O1

=

AB

BO1

height of image
A′B′ = B′O1×

AB

BO1

A'B = 140 ×

100

3000

= 4.7 cm
(c) Now we want to find the height of final image A′′B′′ assuming it to be formed at 25 cm.

Magnification by the eyepiece is
me = (1+

D

fe

) = (1+

25

5

) = 6
Now height of final image, me =

A"B"

A′B′

A′′B′′ = me × A′B′ = 6 × 4.7 = 28.2 cm