A person with a normal near point `(25 cm)` using a compound microscope with an objective of focal length `8.0 mm` and eye piece of focal length `2.5 cm` can bring an object placed `9.0 cm` from the objective in sharp focus. What is the separation between the two lenses ? Calculate the magnifying power of the microscope ?

Options

A. `9.47 cm, 88`
B. `3.36 cm, 44`
C. `6.00 cm, 22`
D. `4.79 cm, 11`

Answer

A

Explanation

Here, `d = 25 cm, f_(o) = 8.0 mm, f_(e) = 2.5 cm, u_(o) = -9.0 mm = -9.0cm`
Now, `(1)/(v_(e)) - (1)/(u_(e)) = (1)/(f_(e)) :. (1)/(u_(e)) = (1)/(v_(e)) - (1)/(f_(e)) = (1)/(-25) - (1)/(2.5) = -(11)/(25)`
`( :' v_(e) = -d = - 25 cm )`
`u_(e) = (-25)/(11) = 2.27 cm`
Again, `(1)/(v_(o)) - (1)/(u_(o)) = (1)/(f_(o))`
`(1)/(v_(o)) = (1)/(f_(o)) + (1)/(u_(o)) = (1)/(0.8) + (0.9-0.8)/(0.72) = (0.1)/(0.72)`
`v_(o) = (0.72)/(0.1) = 7.2 cm`
Therefore, separation between two lenses,
`= u_(e) + v_(o) = 2.27+7.2= 9.47 cm` ltbr Magnifying power, `m = (v_(o))/(u_(o)) (1+(d)/(f_(e))) = (7.2)/(0.9)(1+25/2.5) = 88`