Draw a ray diagram to show the working of a compound microscope. Deduce an expression for the total magnification the final image is formed at the near point.
In a compound microscope , an object is placed at a distance of 1.5 cm from the objective of focal length 1.25cm. if the eye piece has a focal length of 5 cm and the final image is formed at the near point, estimate the magnifying the power of the microscope.

Explanation

Ray diagram for a compound mucroscope

Total angular magnification, `m=(beta)/(alpha)`
`beta rarr` Angle subtended by the image
`alpha rarr` Angle subtended by the object
Since `alpha` and `beta` are small,
`tan alpha` approx `alpha` and `tan beta` approx `beta`
`m=(tan beta)/(tan alpha)`
`tan alpha=(AB)/(D)`
And
`tan beta=(A "B")/(D)`
`m=(tan beta)/(tan alpha)=(A"B")/(D)xx(D)/(AB)=(A"B")/(AB)`
On multiplying the numerator and the denominator with `A'B'`, we obtain
`m=(A"B"xxA'B')/(A'B'xxAB)`
Now, magnification produced by objective, `m_(0)=(A'B')/(AB)`
Magnification produced by eyepiece, `m_(e)=(A"B")/(AB)`
Therefore,
Total magnification, `(m)=m_(0)m_(e)`
"image distance for image produced by objective lens")/("Object distance for the objective lens")`
`m_(e)=(1+(D)/(f_(e)))`
`f_(e) rarr` Focal length of eyepiece
`m=m_(0)m_(e)`
`=(v_(0))/(u_(0))(1+(D)/(f_(e)))`
`v_(0) approx L` (Separation between the lenses)
`u_(0) approx-f_(0)`
`therefore m=(-L)/(f_(0))(1+(D)/(f_(e)))`
`u_(0)=-1.5 cm`
`f_(0)=+1.5 cm`
`(1)/(f_(0))=(1)/(v_(0))-(1)/(u_(0))`
`(1)/(1.25)=(1)/(v_(0))-(1)/(1.5)=(100)/(125)-(10)/(15)=(1500-1250)/(1875)`
`(1)/(v_(0))=(250)/(1875)`
`v_(0)=+7.5 cm`
`f_(e)=+5 cm`
`m=(v_(0))/(u_(0))(1+(D)/(f_(e)))=(7.5)/(-1.5)(1+(25)/(5))=-(7.5)/(-1.5)xx6`
`m=-30`.