Physics, asked by ksssanjay7610, 1 year ago

The magnifying power of an astronomical telescope in the normal adjustment position is 100. The distance between the objective and the eye-piece is 101 cm. Calculate the focal lengths of the objective and of the eyepiece.

Answers

Answered by abhi178
12
it is given that, distance between the objective and the eye-piece , f_0+f_e= 101 cm.....(1)

magnification, m = \frac{f_0}{f_e} = 100

or, f_0=100f_e ......(2)


from equations (1) and (2),

100f_e+f_e=101

or, 101f_e=101\implies f_e=1

and f_0 = 100f_e = 100

hence, focal length of objective = 100cm

focal length of eyepiece = 1cm

Answered by vanshita1981
0

Answer:

100 cm and 1 cm respectively

Explanation:

Similar questions