Question

A small telescope has an objective lens of focal length 144cm andan eyepiece of focal length 6.0cm. What is the magnifying power ofthe telescope? What is the separation between the objective and the eyepiece?

Answer 1

Given that the focal length of the objective lens is 144 cm and focal length of the eyepiece is 6 cm.

Let m be the magnifying power of the telescope, then

m= f o f e

Here, f o be the focal length of the objective lens and f e be the focal length of the eyepiece.

Substitute the values in the above expression.

m= 144 6 =24

Let d be the separation between the objective lens and eyepiece, then

d= f o + f e

Substitute the values in the above expression.

d=144+6 =150 cm

Hence, the value of magnifying power is 24 and the separation between the objective lens and eyepiece is 150 cm.