The separation `L` between the objective `(f = 0.5 cm)` and the eye-piece `(f = 5cm)` of a compound microscope is `7cm`. Where should a small object be placed so that the eye is least strained to see the image? Fin dthe angular magnification produced by the microscope.

Explanation

`f_(0) = 0.5 cm, f_(e) = 5cm, L = 7 cm`
The eye is least strained i.e. final image is formed at infinite.
`L = v_(0) +f_(e)`
`7 = v_(0) +5 rArr v_(0) = 2 cm`
For objective: `u_(0) =- x`
`(1)/(v_(0)) - (1)/(u_(0)) = (1)/(f_(0))`
`(1)/(2) - (1)/(-x) = (1)/(0.5)`
`(1)/(x) = 2-(1)/(2) = (3)/(2)`
`x = (2)/(3)cm, u_(0) =- x =- (2)/(3) cm`
ANgular magnification
`m = (v_(0))/(u_(0)).(D)/(f_(e))`
`= (2)/(-2//3).(25)/(5)`
`=- 3xx5 =- 15`