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 : -