(a) We want the final image at least distance of distinct vision. Let the object in front of objective is at distance u0.
Let us first find the ue, the object distance for eyepiece.
Here ve = - 25 , fe = 6.25 , ue = ?

1

ve

−

1

ue

=

1

fe

1

−25

−

1

ue

=

1

6.25

or −

1

ue

=

1

6.25

+

1

25

=

4+1

25

ue = – 5 cm
So, image distance of objective lens
vo = 15 – ue = 15 – 5 = 10 cm
Now we can get required position of object in point of objective.

1

v0

−

1

u0

=

1

f0

or

1

+10

−

1

ue

=

1

2

1

u0

=

1

10

−

1

2

=

1−5

10

⇒ u0 = - 2.5 cm
So, the object should be 2.5 cm in front of objective lens.
Magnifying power (most strained eye)
m = -

v0

u0

[1+

D

fe

] or m = -

10

−2.5

[1+

25

6.25

] = 4 [5] = 20
(b) We want the final image at infinity. Let us again assume the object in front of objective at distance uo.

Since ve = ∞
∴

1

ve

−

1

ue

=

1

fe

The object distance ue for the eyepiece should be equal to fe = 6.25 cm to obtain final image at ∞.
So, image distance of objective lens
vo = 15 – fe = 15 – 6.25 = 8.75 cm
Now, lens formula,

1

v0

−

1

u0

=

1

f0

⇒

1

8.75

−

1

u0

=

1

2

or

1

u0

=

1

8.75

−

1

2

=

2−8.75

17.5

⇒ u0 = -

17.5

6.75

cm = - 2.59 cm
Magnifying power (most relaxed eye)
m = -

v0

u0

[

D

fe

] ⇒ m = -

8.75

−2.59

[

25

6.25

] = 13.5