A compound microscope consists of an objective lens of focal length 2.0 cm and an eyepiece of focal length 6.25 cm separated by a distance of 15 cm. How far from the objective should an object be placed in order to obtain the final image at (a) the least distance of distinct vision (25cm), and (b) at infinity? What is the magnifying power of the microscope in each case ?

Explanation

`f_(0) cm , f_(e) = 6.25` cm
Distance between object lens and eyepiece = 15 cm
`therefore (1)/(v_(e)) - (1)/(u_(e)) = (1)/(f_(e))`
- `(1)/(u_(e)) = (1)/(f_(e)) - (1)/(v_(e)) = (1)/(6.25) - (1)/((-25)) = (1)/(5) cm " " therefore u_(e) = - 5 ` cm
Object is on the left of object lens.
Distance of the image from the objective lens = 15 -5 = 10 cm
i.e., `v_(0) = 10` cm
`(1)/(v_(0)) - (1)/(u_(0)) = (1)/(f_(0)) " " - (1)/(u_(0)) = (1)/(f_(0)) - (1)/(v_(0)) = (1)/(2) - (1)/(10) = (5 - 1)/(10) = (4)/(10)`
`u_(0) = -2.5` cm
Magnifying power = M = `(v_(0))/(-u_(0)) [ 1 + (D)/(f_(e)) ] = (10)/(+ 2.5) [ 1 + (25)/(6.25) ] ` = 20
b. When final image is at infinity , then object must be placed at the focus of the eyepiece.
`u_(e) = - f_(e) = - 6.25 ` cm
`v_(0) = L - |u_(e)| = 15 - 6.25 = 8.75 ` cm
`(1)/(v_(0)) - (1)/(u_(0)) = (1)/(f_(0)) " " - (1)/(u_(0)) = (1)/(f_(0))- (1)/(v_(0)) = (1)/(2) - (1)/(8.75) = (8.75 - 2)/(8.75 xx 2) = (6.75)/(15.50)`
`u_(e) = - 2.6 ` cm
M = `((-v_(0))/(v_(0))) . (D)/(f_(e)) = (8,75)/(2.6) xx (25)/(6.25) = 13.5`