A compound microscope has a magnifying power of 100 when the image is formed at infinity. The objective has a focal length of 0.5 cm and the tube length is 6.5 cm. Find the fbcal length of the eyepiece.

Explanation

When the final image is at infinity,
`u_(e)=f_(e)=`tube light`-v_(o)`
`therefore f_(e)=6.5-v_(o)….(i)`
Since, `M_(oo)-(v_(o))/(u_(o))cdotD/(f_(e))`
`therefore 100=(v_(o))/(u_(o))cdot25/(f_(e))or(v_(o))/(u_(o)f_(e))=4`….(ii)
For the objective,
`1/(v_(o))-1(-u_(o))=1/(f_(o))=1/1.05rArr1/(v_(o))+1/(u_(o))=2`
we have three unknown `v_(o),u_(o)` and `f_(e)` solving these three equations we get
`f_(e)=2 cm`