Draw a ray diagram to show the working of a compound microscope. Deduce an expression for the total magnification when 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 1.25 cm. If the eye piece has a focal length of 5 cm and the final image is formed at the near point, estimate the magnifying power of the microscope.

#### Solution

**Ray diagram for a compound microscope**

Total angular magnification, `m = beta/alpha`

β → Angle subtended by the image

α → Angle subtended by the object

Since α and β are small,

`tan alpha ≈ alpha and tan beta ≈ beta`

`m= (tanbeta)/(tanalpha)`

`tan alpha = (AB)/D`

And

`tan beta = (A''B'')/D`

`m = (tanbeta)/(tanalpha) = (A''B'')/D xx D/(AB) = (A''B'')/(AB)`

On multiplying the numerator and the denominator with A′B′, we obtain

`m = (A''B'' xx A'B')/(A'B' xx AB)`

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}

`m_0 = ( V_0) / (u_0) =(\text { Image distance for image produced by objective lens})/(\text { Object distance for the objective lens})`

`m_e = (1+D/(f_e))`

*f*_{e }→ Focal length of eyepiece

`m = m_0m_e`

`= V_0/u_0(1+D/f_e)`

`V_0 ≈ L`(Separation between the lenses)

`u_0 ≈ -f_0`

`therefore m = (-L)/(f_0) (1 +D/f_e)`

`u_0 = -1.5 cm`

`f_0 = +1.5cm`

`1/f_0 = 1/v_0 - 1/u_0`

`1/1.25 =1/v_0 + 1/1.5`

`1/v_0=1/1.25 - 1/1.5`

`= 100/125 - 10/15`

`= (1500 -1250)/1875`

`1/v_0 = 250/1875`

`v_0 = + 7.5 cm`

`f_e = + 5cm`

`m =v_0/u_0 (1+D/f_e)`

`= 7.5 / - 1.5 (1+25/5)`

`= - 7.5/1.5 xx 6`

`m =-30`