Magnifying Power of Compound Microscope

We all have used microscopes in our school laboratory

These optical instruments made it possible for us to see and study tissues and cells of plants

In order to study the cells, a simple microscope would not suffice with it's limited magnification of $40 X$

To magnify beyond $40 X$ , a second lens is added to a simple microscope. This set up is called a compound microscope

The two lenses in a compound microscope are called Objective and Eyepiece lens

The lens through which the final image is observed is called the eyepiece lens whereas the lens just above the object is called the objective

The combination of lenses in a compound microscope allows magnification in the range of $40 \times \to 1000 \times$

Let us understand the working principle of a compound microscope

The object $A B$ is placed outside the focus of the objective lens $(F_{1}$ ). The objective lens forms a real, magnified and inverted image $A_{1} B_{1}$

The image $A_{1} B_{1}$ becomes the object for the second lens, the eyepiece

The distance between the two lenses is adjusted such that the image $A_{1} B_{1}$ lies within the focus of the eyepiece, $F_{2}$

The eyepiece forms a magnified, virtual and inverted (as compared to the original object) image

Hence together, the lenses in a compound microscope produce a highly magnified, virtual and inverted image of the object

Let us see how the formula for magnification by a compound microscope can be obtained

We know the magnifying power ( $M$ ) is defined as - $M = \frac{β}{α}$

$β$ is the angle subtended by the final magnified image on the eye; whereas $α$ is the angle subtended by the original object on the eye when placed at the least distance of distinct vision

From the ray-diagram, we can see that $β = \frac{A _{1} B _{1}}{u _{e}} = \frac{A _{2} B _{2}}{v _{e}}$

From the figure, $α = \frac{A B}{D}$

We can then calculate the magnifying power, $M = \frac{β}{α}$

Re-arranging the terms $M = \frac{A _{1} B _{1}}{A B} \times \frac{D}{u _{e}}$

$We know, A_{1} B_{1} = v_{o} A B = u_{o}$ $∴ M = \frac{v _{o}}{- u _{o}} \times \frac{D}{u _{e}} -ve sign is used for u_{o} according to the sign-convention$

Therefore we can write $M = - \frac{v _{o}}{u _{o}} \times \frac{D}{u _{e}}$

Applying the lens formula on the eyepiece, $\frac{1}{f _{e}} = \frac{1}{v _{e}} - \frac{1}{u _{e}}$

According to the sign-convention $v_{e}$ and $u_{e}$ will be negative

Therefore the lens formula becomes $\frac{1}{f _{e}} = \frac{1}{- v _{e}} - \frac{1}{- u _{e}} \to \frac{1}{u _{e}} = \frac{1}{f _{e}} + \frac{1}{v _{e}}$

Using the value of $\frac{1}{u _{e}} we can write M = - \frac{v _{o}}{u _{o}} (\frac{D}{f _{e}} + \frac{D}{v _{e}})$

Now there can be two cases regarding the final image

When the final image is at D, $v_{e} = D$ $M = - \frac{v _{o}}{u _{o}} (1 + \frac{D}{f _{e}})$

M, in this case, can also be written as $M = \frac{f _{o}}{u _{o} - f _{o}} (1 + \frac{D}{f _{e}})$

When the final image is at infinity, $v_{e} = \infty$ $M = - \frac{v _{o}}{u _{o}} \times \frac{D}{f _{e}}$

Revision

The magnifying power (M) of a compound microscope when the final image is formed at D is given by, $M = - \frac{v _{o}}{u _{o}} (1 + \frac{D}{f _{e}})$

The magnifying power (M) of a compound microscope when the final image is formed at D is also given by, $M = \frac{f _{o}}{u _{o} - f _{o}} (1 + \frac{D}{f _{e}})$

The magnifying power (M) of a compound microscope when the final image is formed at infinity is given by, $M = - \frac{v _{o}}{u _{o}} \times \frac{D}{f _{e}}$

The End