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(a) A giant refracting telescope at an observatory has an objective lens of focal length 15m. If an eyepiece of focal length 1.0cm is used, what is the angular magnification of the telescope? (b) If this telescope is used to view the moon, what is the diameter of the image of the moon formed by the objective lens? The diameter of the moon is 3.48 × 106m, and the radius of lunar orbit is 3.8 × 108m.
By
1. Given Data
- Focal length of Objective (fo): 15 m = 1500 cm
- Focal length of Eyepiece (fe): 1.0 cm
- Diameter of Moon (D): 3.48 × 106 m
- Radius of Lunar Orbit (Distance u): 3.8 × 108 m
2. Part (a): Angular Magnification
The angular magnification (M) of a telescope in normal adjustment is given by:
M = fo / fe
Substituting the values (ensuring both are in cm):
M = 1500 / 1.0
M = 1500
Angular Magnification = 1500
3. Part (b): Diameter of the Image
The image of the moon is formed by the objective lens at its focal plane. The angle subtended by the moon at the objective is equal to the angle subtended by the image.
Step A: Angle subtended by Moon (α)
α = Diameter of Moon / Distance from Earth
α = (3.48 × 106) / (3.8 × 108)
α ≈ 0.009158 radians
Step B: Size of Image (d)
The size of the image (d) formed at the focal length (fo) is:
d = fo × α
d = 15 m × 0.009158
d = 0.13737 m
Converting to centimeters:
d ≈ 13.74 cm
Final Answer
- (a) Angular Magnification: 1500
- (b) Image Diameter: 13.74 cm
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