A compound microscope consists of an object lens of focal length 2 cm and an eyepiece of focal length 5 cm. When an object is placed at 2.2 cm from the object lens, the final image is at 25 cm from the eye lens. What is the distance between the lenses ? What is the total linear magnification ?

Explanation

Given that `f_(0)=2, f_(e)=5, u_(0)=2.2,`
D = 25 cm
`(1)/(f_(0))=(1)/(u_(0))+(1)/(v_(0))`
`(1)/(v_(0))=(1)/(f_(0))-(1)/(u_(0))=(1)/(2)-(1)/(2.2) rArr (1)/(v_(0))=(2.2-2)/(2xx2.2)`
`v_(0)=22cm`
For the eye-piece, the distance of the image `v_(e)=25cm`
`(1)/(f_(e))=(1)/(u_(e))-(1)/(v_(e))" (For virtual image)"`
`(1)/(u_(e))=(1)/(f_(e))+(1)/(v_(e))=(1)/(5)+(1)/(25)`
`rArr (1)/(uu_(e))=(25+5)/(5xx25)`
`u_(e)=4.166`
i) The distance between the two lenses
`L=v_(0)+u_(e)`
`L=22+4.166`
`L = 26.166cm`
ii) Magnification `(m)=(v_(0))/(u_(0))(1+(D)/(f_(e)))`
`m=(22)/(2.2)(1+(25)/(5))`
`m=10xx6`
`m=60`