A compound microscope consists of an objects lens of focal length 2.0 cm and an eyepiece of focal length 20cm? If the system a converging or a diverging lens? Ignore thickness of the lenses.

Explanation

Focal length of the objective lens, f1 = 2.0cm
Focal length of the eyepiece, f2=6.25 cm
Distance between the objective lens and the eyepiece, d=15cm
a) Least distance of distinct vision, `d^(') = 25cm`
Object distance for the eyepiece = u2
According to the lens formula, we have the relation:
`1/v_(2) -1/u_(2)= 1/f_(2)`
`1/u-(2)=1/v_(2)-1/f_(2)`
`=1/-25 -1/6.25 = (-1-4)/(25) = -5/25`
`therefore u_(2)=5 cm`
Image distance for the obective lens, `v_(1)=d+u_(2)=p15-5=10cm`
Object distance for the objective lens = u1
According to the lens formula, we have the relation:
`1/v_(i) -1/u_(1)=1/f_(1)`
`1/u_(1)=1/v_(1)-1/f_(1)`
`=1/10-1/2=(1-5)/10=-4/10`
`therefore u_(i) = -2.5cm`
Magnitude of the object distance, `|mu_(1)|`= 2.5 cm
The magnifying power of a compound microscope is given by the relation:
`m=v_(i)/|mu_(1)|(1+25/6.25)= 4(1+4)=20`
Hence, the magnifying power of the microscope is 20.
the final image is formed at infinity.
Image distance for the everyone , `v_(2)= infty`
Object distance for the eyepiece = u2
According to the lens formula, we have the relation:
`1/v_(2)-1/u_(2)=1/f_(2)`
`1/infty-1/mu_(2)=1/6.25`
`therefore mu_(2) = -6.25 cm`
Magnitude of the object distance, `|mu_(1)|=2.59cm`
The magnifying power of a compound is given by the relation:
`m=v_(i)/(|mu_(1)|)(d^(')/|mu_(2)|)`
`=8.75/2.59 xx 25/6.25 = 13.51`
Hence, the magnifying power of the microscope is 13.51