Question

When a telescope is adjusted for normal vision the distance of the objective from the eyepiece is found to be 80 CM the magnifying power of the telescope is 19 what are the focal lengths of the lenses

Answer 1

The focal length of object is 76 cm

The focal length of eyepiece is 4 cm  

Explanation:

Given as :

A telescope is adjusted for normal vision

The magnification power of the telescope = M = 19

The distance of the objective from the eyepiece = 80 cm

Let The focal length of object = $F_o$

Let The focal length of eyepiece = $F_e$

According to question

∵  magnification power = $\dfrac{focal length of object}{focal length of eyepiece}$

Or,                               M = $\dfrac{F_o}{F_e}$

Or,                               19 = $\dfrac{F_o}{F_e}$

∴                                  $F_o$  = 19 $F_e$                      ..........1

Again

The distance of the objective from the eyepiece = 80 cm

i.e   $F_o$ + $F_e$  = 80                            ...........2

Solving eq 1 and eq 2

      19 $F_e$ + $F_e$  = 80  

Or,       20  $F_e$  = 80

∴                $F_e$ = $\dfrac{80}{20}$

i.e              $F_e$ = 4  cm

Now, Put the value of $F_e$ in eq 1

So, ∵  $F_o$  = 19 $F_e$

Or,     $F_o$  = 19 × 4 cm

∴        $F_o$  = 76 cm

So, The focal length of object = $F_o$ = 76 cm

And The focal length of eyepiece = $F_e$ = 4 cm

Hence, The focal length of object is 76 cm

And The focal length of eyepiece is 4 cm  Answer