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.

f₀= focal length of the object

f_e= focal length of the eyepiece

D= least distance of the distinct vision

(2) Given parameters

f₀= 150 cm

f_e= 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/f_e= 1/v_e – 1/u_e

1/5 = 1/(-25) – 1/u_e

u_e = -25/6

agnefication due to eyepiece

m_e = -25/(-25/6)

m_e = 6 cm

Total magnification (M) = m_e × m₀

M = 6 × 5 × 10^-4

Size of the final image = 30cm