Question

A small telescope has an objective lens of focal length 144 cm and an eyepiece of focal length 6.0 cm. What is the magnitude of magnifying power of the telescope? What is the separation between the objective and the eyepiece? (Here final image formed at infinity)​

Answer 1

Answer:

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Answer 2

Given:

Objective's focal length,

$f_{0}=144 \ cm$

Eyepiece's focal length,

$f_{e}=6.0 \ cm$

To Find:

The separation between the eyepiece & objective = ?

Solution:

As we know,

The magnifying power of a telescope will be:

⇒  $m=\frac{-f_{0}}{f_{e}}$

On putting the estimated values, we get

⇒      $=\frac{-144}{6.0}$

⇒      $=-24$

Now,

Separation between them will be:

⇒  $L=f_{0}+f_{e}$

⇒      $=144+6.0$

⇒      $=150 \ cm$

So that the separation value between the objective lens & eyepiece will be "150 cm".