Final answer:
The angular magnification of the telescope is 50x. The angle subtended by the 25,000 km diameter sunspot is 0.036°. The angle of the telescopic image is 0.018°.
Step-by-step explanation:
(a) Angular magnification:
The angular magnification of a telescope is given by the formula:
M = (D/fe)/(f0/fe - 1)
where D is the diameter of the objective lens, fe is the focal length of the eyepiece, and f0 is the focal length of the objective lens.
Plugging in the given values, we have:
D = 2.00 m
fe = 4.00 cm = 0.04 m
f0 = 2.00 m
Using these values in the formula, we get:
M = (2.00/0.04)/(2.00/0.04 - 1) = 50
Therefore, the telescope's angular magnification is 50x.
(b) Angle subtended by the sunspot:
The angle subtended by a sunspot can be calculated using the formula:
θ = D/R
where D is the diameter and R is the distance.
Plugging in the given values, we have:
D = 25,000 km = 25,000 x 10^3 m
R = radius of curvature = 2.00 m
Using these values in the formula, we get:
θ = (25,000 x 10^3/2.00) = 0.036°
Therefore, the angle subtended by the sunspot is 0.036°.
(c) Angle of the telescopic image:
The angle of the telescopic image can be calculated using the formula:
θ' = Mθ
where M is the angular magnification and θ is the angle subtended by the sunspot.
Using the values calculated in parts (a) and (b), we have:
θ' = 50 x 0.036° = 0.018°
Therefore, the angle of the telescopic image is 0.018°.