Question

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`.

Answer 1

Here, `d = 150 mm = 150 xx 10^(-3) m`,
`f_(0) = 4.0m, f_(6) = 25.0 mm =25 xx 10^(-3)m`
`lambda = 6000 Å = 6 xx 10^(-7) m`
if we assume that final image is formed at infinity, then magnifying power,
`m = (f_(0))/(f_(e)) = (4.0)/(25 xx 10^(-3)) = 160`
Resolving power `= (D)/(1.22 lambda) = (150 xx 10^(-3))/(1.22 xx 6 xx 10^(-7))`
`= 2.05 xx 10^(5)`
Disatnce between objective and eye piece
`= f_(0) + f_(e) + 4.0 + 25 xx 10^(-3) = 4.025 m`