(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.
a)
Given: The focal length of objective lens is 15 m , and focal length of the eyepiece is 1 cm .
The angular magnification can be calculated as,
α = f o f e
where, the focal length of objective lens is f o and focal length of the eyepiece is f e .
By substituting the given values in the above equation, we get,
α = 15 m 1 cm = 15 10 − 2 = 1500
Thus, the angular magnification is 1500 .
b)
Given: The diameter of moon is 3.48 × 10 6 m , and the radius of lunar orbit is 3.8 × 10 8 m .
The angle subtended is given as,
θ = d m r l
where, the diameter of the moon is d m and the radius of lunar orbit is r l .
By substituting the given values in the above equation, we get,
θ = 3.48 × 10 6 3.8 × 10 8 = 0.00915 radians (1)
The angle subtended by the image of diameter of the moon is given as,
θ ′ = d m ′ f o
where, the diameter of image of the moon is d m ′ and the focal length of objective lens is f o .
By substituting the given values in the above equation, we get,
θ ′ = d m ′ 15 m (2)
As telescope is used to view the moon; hence, the angle subtended by the image would be same as the angle subtended by the diameter of the moon therefore,
θ = θ ′
From equation (1) and (2),
0.00915 = d m ′ 15 m d m ′ = 15 × 0.00915 ≈ 13.7 × 10 − 2 m ≈ 13 .7 cm
Thus, the diameter of the image formed by the objective lens is 13 .7 cm .
a)
Given: The focal length of objective lens is
The angular magnification can be calculated as,
where, the focal length of objective lens is
By substituting the given values in the above equation, we get,
Thus, the angular magnification is
b)
Given: The diameter of moon is
The angle subtended is given as,
where, the diameter of the moon is
By substituting the given values in the above equation, we get,
The angle subtended by the image of diameter of the moon is given as,
where, the diameter of image of the moon is
By substituting the given values in the above equation, we get,
As telescope is used to view the moon; hence, the angle subtended by the image would be same as the angle subtended by the diameter of the moon therefore,
From equation (1) and (2),
Thus, the diameter of the image formed by the objective lens is
