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? ​

Answer 1

AnSwer :

Focal length of the objective lens, f${\sf{_o}}$ = 144cm.

Focal length of the eyepiece, f${\sf{_e}}$ = 6.0cm

The magnifying power of the telescope is given as :-

➠ m = f${\sf{_o}}$/f${\sf{_e}}$

➠ m = 144/6

➠ m = 24

The separation between the objective lens and the eyepiece is calculated as :-

➠ L = f${\sf{_o}}$ + f${\sf{_e}}$

➠ L = 144 + 6

➠ L = 150cm.

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

Answer 2

Solution :-

As per the given data ,

• The focal length of objective lens (fo) = 144 cm

• The focal length of eyepiece (fe) = 6 cm

The magnifying power of the telescope is given by ,

➜  m = fo / fe

➜  m = 144 / 6

➜  m = 24

The magnifying power of the telescope is 24

The separation between the objective and the eyepiece

= fo + fe

= 144 + 6

= 150 cm

The separation between the objective and the eyepiece(under normal adjustment )  is 150 cm