The objective of an astronomical telescope has a diameter of 150 mm and a focal length of 4.00 m.The eyepiece has a focal length of 25.00 mm. Calculate the magnifying and resolving power of telescope.(λ=6000 A for yellow colour).

the magnification is given as

m = fo / fe

here

fo = focal length of objective = 4m

fe =  focal length of eyepiece = 25mm = 0.025m

so,

m = 4 / 0.025

thus, we have

magnification

m = 160

.

the resolving power is given as

R = 1.22λ / D

here

λ  = wavelength of light = 6000 A = 6000 x 10-10 m

D = diameter of the objective = 150mm = 1.5m

so,

R = (1.22 x 6000 x 10-10) / 1.5

thus, we have

resolving power

R =  4800 A 

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