1 ) Focal length of objective is 1.5 cm . What should be the focal length of eyepiece to get the magnification of 25 ? A ) 0.06 cm , B ) 1.09 cm , C ) 2.6 cm , D ) None of these ,

1 )  Focal length of objective is 1.5 cm . What should be the focal length of eyepiece to get the magnification of 25 ?

A ) 0.06 cm ,                                    B ) 1.09 cm ,

C ) 2.6 cm ,                                      D ) None of these ,

Correct Answer : -   A ) 0.06 cm ,

Solution : -

We know formula for magnifying that : -

M = fo / fe . . . . . . . . . eq. 1

M = magnifying power  =  25 ,

fo = 1.5 cm ,

fe = ?

So put in eq. 1 ;

25 = 1.5 / fe .

fe = 1.5 / 25 = 0.06 cm ,  Answer .

 

2 ) The focal length of eye piece and objective of a compound microscope are 5 cm and 1 cm respectively and the length of tube is 20 cm .  Calculate the magnifying power of microscope when the final image is at infinity .  The least distance of distinct vision is 25 cm ?

A ) 40 ,                                             B ) 70 ,

C ) 90 ,                                            D ) None of these ,

Correct Answer : -   B ) 70 ,

Solution : -

So ,  Ve = ∞ ,  ( final image )  

So , length of microscope : - 

L = Vo + fe ,

Vo = L - fe = 20 – 5 = 15 cm ,   And 

∴  fo = 1 cm .

So ,  1 / u˳ = (1 / fo) – (1 / V)  

1 / u˳ =  (1 / 1) – (1 / 25)  =  + 14 / 15 .

So ,   u˳ =  + 15 / 14  .

Now we can find the magnifying power by formula : -

m = ( V˳ / u˳ ) ( d / fe )  =  { 15 / (15/14) } ( 25 / 5 )  ,

 m = 70 ,   Answer .

 

3 )  Power of lens : -  P = 1 / f .

 

4 ) Magnifying power of telescope : -

m =  fo / fe .

 

5 )  Formulas : -