Question

Explain simple microscope on the basis of the following points:
(a) Ray diagram for the formation of image;
(b) Expression for magnifying power when the final image is formed:
(i) At the least distance of distinct vision;
(ii) At infinity.

Answer 1

(a) Ray diagram : - (Ref. image).

(b) Expression for magnifying power: The ratio of the visual angle of image to the visual angle of object, placed at the least distance of distinct vision, is called magnifying power.
$m=\frac{\text{Visual angle of image}}{\text{Visual angle of object placed at the least distance of distinct vision}}$
or $m=\frac{\beta }{\alpha }$
As the angles $\alpha$ and $\beta$ are small.
Therefore, $m=\frac{\tan \beta }{\tan \alpha }$ ......(i)
In $\Delta AOB$, $\tan \beta =\frac{AB}{OB}$
and in $\Delta A^{{}^{\prime \prime }}O{B}^{{}^{\prime }}$,
$\tan \alpha =\frac{A^{{}^{\prime \prime }}{B}^{{}^{\prime }}}{B^{{}^{\prime }}O}$
$\therefore$ $m=\frac{AB/OB}{A^{{}^{\prime \prime }}{B}^{{}^{\prime }}/B^{{}^{\prime }}O}$
or $m=\frac{AB}{OB}\times \frac{D}{AB}$
($\because A^{{}^{\prime \prime }}{B}^{{}^{\prime }}=AB$ and $BO=D=$Least distance of distinct vision)
Or $m=\frac{D}{OB}$
By sign convention, $D$ will be negative and $OB=-u$.
$\therefore$ $m=\frac{-D}{-u}$
Or $m=\frac{D}{u}$ .....(i)
Where, $u$ is the distance of object from the lens.
(i) when the image is formed distinct vision at the least distance of:
from lens formula, $\frac{1}{f}=\frac{1}{v}-\frac{1}{u}$
$\frac{1}{f}=\frac{1}{-v}-\frac{1}{-u}$ (from sign convention)
Or $\frac{1}{f}=\frac{-1}{v}+\frac{1}{u}$
But $v=D=$ Least distance of distinct vision
$\therefore$ $\frac{1}{f}=\frac{-1}{D}+\frac{1}{u}$
Or $\frac{D}{f}=-1+\frac{D}{u}$
Or $1+\frac{D}{f}=\frac{D}{u}$
Putting the value in equation (ii)
$m=1+\frac{D}{f}$
(ii) When the image is formed at infinity:
$u=f$
From equation (i) $m=\frac{D}{f}$
For greater magnification, focal length should be small.