The objective of an astronomical telescope has a diameter of 150 mm and a focal length of 4 m. The eyepiece has a focal length of 25 mm. Calculate the magnifying and resolving power of telescope

# The objective of an astronomical telescope

**prasanna**#2

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 ${{10}^{6}}$