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

Explanation

`F_(0) = 8 mm = 0.8, f_(e) = 2.5 `cm
D = 25 cm , `v_(e) - D = - 25 cm, u_(e) = - 9 mm = 9.0 `cm
`m_(e) = (v_(e))/(u_(e)) = 1 + (D)/(f_(e)) " " (-25)/(u_(e)) = 1 + (25)/(2.5) = 11 " " therefore u_(e) = (-25)/(11) cm`
`(1)/(v_(0)) - (1)/(u_(0)) = (1)/(f_(0))`
`(1)/(v_(0)) = (1)/(f_(0)) + (1)/(u_(0)) = (1)/(0.8) + (1)/(-0.9) " " therefore v_(0) = 7.2 ` cm
Separation between lenses = `|u_(e)| + |v_(e)| = (25)/(11) + 7.2 = 9.47 cm `
`M = (v_(0))/(-u_(0)) [ 1 + (D)/(f_(e)) ]= (72)/(0.9) (1 + (25)/(2.5) ) = 88`