Question

A small telescope has an objective lens of focal length 150 cm and eye piece of focal length 5 cm if this telescope is used to view a 100 m high tower 3 km away, find the height of final image when it is formed 25 cm away from the eye piece . Please answer it quickly. ...tomorrow my exam is there .....best answer will be marked as brainliest. .....

Answer 1

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Answer 2

Hey dear,

● Answer-
h2 = 30 cm

● Explaination-
# Given-
fo = 150 cm = 1.5 m
fe = 5 cm = 0.05 m
h1 = 100 m

# Solution-
Using lens formula for objective,
1/fo = 1/vo - 1/uo
1/1.5 = 1/vo - 1/(-3×10^3)
vo = 1.5 m

Magnification by objective is -
Mo = vo/uo = 1.5 / 3×10^3
Mo = 5×10^-4

Using lens formula for eyepiece,
1/fe = 1/ve - 1/ue
1/0.05 = 1/(-0.25) - 1/ue
ve = -0.042 m

Magnification by eyepiece,
Me = ve/ue = 0.25 / 0.042
Me = 6

Total magnification is-
M = Mo × Me
M = 5×10^-4 × 6
M = 3×10^-3

Height of final image,
M = h2/h1
h2 = M × h1
h2 = 3×10^-3 × 100
h2 = 0.3 m
h2 = 30 cm

Therefore, height of image is 30 cm.

Hope this helps...