TG EAMCET 2 May 2025 Shift 2 Paper

Given, the focal length of the objective lens is f0=1.25cm
The focal length of the eyepiece is
fe=5cm
The separation between the objective and eyepiece is L=7.5cm
The distance of this image from the objective lens is
‌v0=L−fe
‌v0=7.5cm−5cm=2.5cm
The lens formula for the objective lens
‌

1

f0

‌=‌

1

v0

−‌

1

u0

‌

1

u0

‌=‌

1

2.5cm

−‌

1

1.25cm

u0‌=−2.5cm
The magnification of the objective lens is
M0=−‌

v0

u0

=‌

−2.5cm

−2.5cm

=1
The magnification of the eyepiece when the final image is at infinity is Me=‌

D

fe

Me=‌

25cm

5cm

=5
The total magnification is M=M0×Me
M=1×5=5