The objective of an astronomical telescope has the diameter of 150mm and the focal length of 4m.The eye piece has focal length of 25mm.Calculate the magnifying and resolving power of the telescope. What is the distance between the objective and the eye piece. Take wavelength 6000a°
Answers
Explanation:
Given The objective of an astronomical telescope has the diameter of 150 mm and the focal length of 4 m.The eye piece has focal length of 25 mm.Calculate the magnifying and resolving power of the telescope. What is the distance between the objective and the eye piece. Take wavelength 6000 a°
- Given diameter d = 150 mm = 150 x 10^-3 m, fo = 4.0 m, fe = 25 mm = 25 x 10^-3 m, λ = 6000 = 6000 x 10^-7 m
- Now resolving power of telescope = d / 1.22 λ
- = 150 x 10^-3 / 1.22 x 6 x 10^-7
- = 2.05 x 10^5
- Now magnifying power m = focal length of objective / focal length of eye piece = fo / fe
- = 4 / 25 x 10^-3
- = 160
- Now distance between objective and eyepiece will be = fo + fe
- = 4.0 + 25 x 10^-3
- = 4.025 m
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Answer:
Explanation:
The diameter of objective of the telescope
=150 x ${{10}^{-3}}$ m
$f_{ o }$ = 4m
$f_{ e }$ = 25 x ${{10}^{-3}}$ m and
D = 0.25m
Magnifying power, m = $f_{ o }$/$f_{ e }$ (1 + D /$f_{ e }$)
m = -4/25 x ${{10}^{-3}}$(1+ 0.25/25 x ${{10}^{-3}}$
m= -1760
Resolving power = 1/dθ = 1.22λ/D = 1.22 x 6x${{10}^{-7}}$/0.25
=2.9 x ${{10}^{-6}}$ rad
Resolving power = 1/2.9 x ${{10}^{-6}}$
=0.34 x ${{10}^{6}}$