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