Question

Draw a neat labelled diagram of a compound microscope and explain its working. Derive an expression for its magnification.

Accepted Answer

Compound microscope consists of two convex lenses separated by a distance. The lens near the object is called objective and the lens near the eye is called eye piece. The objective lens has small focal length and eye piece has of larger focal length.

The object placed for the objective lens produces an image and this image acts as object for the eye piece lens. The position of the eye piece lens is adjusted such that it produces virtual and magnified image for the eye piece lens.

The ratio of the angle subtended by the final image at the eye when formed at near point to the angle subtended by the object at the eye when imagined to be at near point is called magnifying power.
Imagining that the eye is at the optic centre, the angle subtended by the final image is $α$ . When the object is imagined to be taken at near point it is represented by $I J'$ and $O J$ The angle made by $I J'$ at the eye is $β$ . Then by the definition of magnifying power

$m = \frac{α}{β} = \frac{t a n α}{t a n β}$

$\Rightarrow \frac{I G / I O^{''}}{I J^{'} / I O^{''}} = \frac{I G}{I J^{'}} = \frac{I G}{O J} \cdot (∵ I J^{'} = O J)$

Dividing and multiplying by $I_{1} G_{1}$ on the right side, we get

$m = (\frac{I G}{I_{1} G_{1}}) (\frac{I_{1} G_{1}}{O J})$

Magnifying power of the objective $(m_{0}) = I_{1} G_{1} / O J$

Magnifying power of the eye piece $(m_{e}) = I G / I_{1} G_{1}$

$m = m_{0} \times m_{e}$

From similar triangles $O J O^{'}$ and $I_{1} G_{1} O^{'}$

$\frac{I_{1} G_{1}}{O J} = \frac{O^{'} I_{1}}{O^{'} O}$

$m_{0} = \frac{v_{0}}{u}$ (where, $v_{0}$ is the image distance for objective lens and $u$ is the object distance)

Now the image of objective lens acts as object for the eyepiece lens. Therefore, it works like simple microscope.

$m_{e} = (1 + \frac{D}{f_{e}})$ (Where, $f_{e}$ is the focal length of the eye lens)

Substituting $m_{o}$ and $m_{e}$ in equation.

$m = \frac{v_{0}}{u} (1 + \frac{D}{f_{e}})$

When the object is very close to the $f_{o}$ then,

$u = - f_{0}$ and $v_{0} = L$ (Where $L$ is the length of the microscope)

$m = - \frac{L}{f_{0}} (1 + \frac{D}{f_{e}})$

Draw a neat labelled diagram of a compound microscope and explain its working. Derive an expression for its magnification.

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