Define magnifying power of a telescope. Write its expression.A small telescope has an objective lens of focal length 150cm and an eye piece of focal length 5cm. If this telescope is used to view a 100m high tower 3km away, find the height of the final image when it is formed 25cm away from the eye piece.

Answer:

(1) The magnifying power of the telescope is the ratio of the angle subtended at the eye by the image to the angle subtended at the unaided eye by the object.

\(Magnifying power (M) = \frac{f_{0}}{f_{e}}\left ( 1+\frac{f_{e}}{D} \right )\)

f0= focal length of the object

fe= focal length of the eyepiece

D= least distance of the distinct vision

(2) Given parameters

f0= 150 cm

fe= 5 cm

D= 25 cm

Using the lens equation for an object lens

1/f0 = 1/v0 – 1/u0

1/150 = 1/v0 – 1/(-3 × 105)

v0 = 150cm

Magnefication due to the object lens

m0 = v0/u0

m0 = 150 × 10 -2/3000

m0 =0.05 10 -2

Using the lens equation for an eye-piece

1/fe= 1/ve – 1/ue

1/5 = 1/(-25) – 1/ue

ue = -25/6

agnefication due to eyepiece

me = -25/(-25/6)

me = 6 cm

Total magnification (M) = me × m0

M = 6 × 5 × 10-4

Size of the final image = 30cm

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