Question

Question No. 16.
The magnifying power of an astronomical telescope in normal adjustment is 100. What is the focal length
of the objectives and eyepiece of the distance between them is 101 cm?
A) 1 cm and 10 cm respectively
B) 1 cm and 100 cm respectively
C) 10 cm and I cm respectively
D) 100 cm and I cm respectively

​

Answer 1

Answer:

hi please mark me as brainlist

Explanation:

Magnifying power for normal adjustment M

1=10= fe

fo

...(1)

Also f

o+ff =110cm ..(2)

By solving (1) and (2) we get,

fo = 100cm and fe =1 0cm

Magnifying power when image is formed at least distance of distinct vision

M2 = −fo (D1 + Fe 1 )

Putting the values we get,

M2 = −100-251 + 101

M2

=−14 is the magnifying power of the telescope when image is formed at the least distance vision for normal eye.

.

.

hope you like it

please mark me as brainlist ....