The focal lengths of the two lenses of a compound microscope are 0.5 cm and 1 cm respectively . An object is placed at a distance of 1 cm from the objective. If the final image of the object is formed at a distance of 25 cm from the eye , what is the distance between the two lenses and the magnifying power of the microscope ?

Explanation

In the compound microscope the focal length of the eyepiece is greater than that of the objective. So `f_o=0.5 cm and f_e=1 cm`. For the objective u=-1 cm
If v be tha image distance , then according to the equation of the lens,
`1/v+1/1=1/(0.5) or, 1/v=-1+2=1 or, v=1 cm `
So, the image is formed on the other side of the objective at a distance of 1 cm .
For the eyepiece , `v_e=-25 cm `(the image is virtual and formed at near point ) and `f_e=1 cm `
we have from the equation of the lens ,
`-1/(25)-1/(u_e)=1/1 or, 1/(u_e)=1/(-25)-1 =-(26)/(25) or, u_e=-(25)/(26)=-0.96 cm`
`therefore |u_e|=0.96 cm `
`therefore` The distance between the two lenses
=|v|+|`u_e|=1+0.96=1.96cm`
The magnifying power of the microscope ,
`m=v/u(1+D/(f_e))=1/1(1+(25)/1)=26`