The objective of an astronomical telescope has a diameter of `150 mm` and focal length of `4.0 m`. The eye piece has a focal length of `25.0 mm`. Calculate the magnifying power and resolving power of telescope. What is the distance between objective and eye piece ? Take `lambda = 6000Å`.
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To solve the problem step by step, we will calculate the magnifying power, resolving power, and the distance between the objective and the eyepiece of the astronomical telescope.
### Given Data:
- Diameter of the objective (D) = 150 mm = 150 x 10^-3 m
- Focal length of the objective (f_o) = 4.0 m
- Focal length of the eyepiece (f_e) = 25.0 mm = 25 x 10^-3 m
- Wavelength (λ) = 6000 Å = 6000 x 10^-10 m
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