The separation between the objective and the eyepiece of a compound microscope can be adjusted between 9.8 cm to 11.8 cm. If the focal lengths of the objective and the eyepiece are 1.0 cm and 6 cm respectively, find the range of the magnifying power if the image is always needed at 24 cm from the eye.

Explanation

For eye-piece, it is given that
`f_(e)=6 cm, v_(e)=-24 cm`
`therefore 1/(-24)-1/(-u_(e))=1/6`
`therefore u_(e)=4.8 cm`
When, L=9.8 cm, then `v_(0)`=(9.8-4.8)cm=5.0 cm
`1/5.0-1/(-u_(0))=1/1.0`
`therefore u_(0)=1.25 cm`
on magnifying power,`M=(v_(0)/u_(0))D/(u_(e))=(5.0/1.25)(25.0/4.8)=20.83`
When,L=11.8 cm,then`v_(0)=(11.8-4.8)`cm=7.0 cm
`therefore 1/7.0-1/(-u_(o))=1/1.0oru_(0)=1.17 cm`
`therefore M=(v_(0))/(u_(0))cdotD/(f_(e))=(7.0/1.17)(25.0/4.8)=31.16`