Question

The objective lens and the eyepiece of a telescope are spaced 85 cm apart. If the eyepiece is123 D what is the total magnification of the telescope?

Answer 1

Answer:

The magnification would be "103.55". A further explanation is given below.

Explanation:

The given values are:

Distance between lens and eyepiece,

L = 85 cm

Eyepiece is,

= 123 D

Now,

The refractive power of eye piece will be:

⇒ $\frac{1}{f_e}=123D$

   $f_e=\frac{1}{123D}$

   $f_e=0.813 \ cm$

The length of the telescope will be:

⇒ $L=f_0+f_e$

⇒ $f_0=L-f_e$

On substituting the values, we get

⇒     $=85-0.813$

⇒     $=84.187 \ cm$

Now,

The magnification of the telescope will be:

⇒ $M=\frac{f_0}{f_e}$

⇒      $=\frac{84.187}{0.813}$

⇒      $=103.55$