Question

Describe the compound microscope on the following headings:
(i) Labelled ray diagram of formation of image.
(ii) Magnifying power when final image is formed at least distance of distinct vision.

Answer 1

Where $O\to$ Objective lens, $E\to$ Eye lens, $AB\to$ Object, $PQ\to$ Image.
(ii) Calculation of magnification: Ratio of angle subtended by an image to the angle subtended by an object placed at least distance of distinct vision is called magnifying power of microscope.
If angle subtended by an image is $\beta$ and angle by an object placed at $D$ is given by $\alpha$. Therefore, magnifying power
$m=\frac{\beta }{\alpha }$ .......(i)
$\therefore \alpha =\frac{AB}{D}$
Where, $D\to$ Least distance of distinct vision.
As eye is close to the eye lens, we can take the angle subtended at lens for human eye also.
$\therefore$ Angle made by an image $PQ$
$\beta =\frac{PQ}{EQ}=\frac{A^{{}^{\prime }}{B}^{{}^{\prime }}}{E{B}^{{}^{\prime }}}$
or $\beta =\frac{A^{{}^{\prime }}{B}^{{}^{\prime }}}{u_c}$
where, $u_c\to E{B}^{{}^{\prime }}\to$ Distance of image from eye lens
Substituting above values in equation (i),
$m=\frac{\frac{A^{{}^{\prime }}{B}^{{}^{\prime }}}{u_e}}{\frac{AB}{D}}=\frac{A^{{}^{\prime }}{B}^{{}^{\prime }}}{AB}\times \frac{D}{u_e}$ .......(ii)
Now, in similar triangle $\Delta A^{{}^{\prime }}O{B}^{{}^{\prime }}$ and $\Delta AOB$
$\frac{A^{{}^{\prime }}{B}^{{}^{\prime }}}{AB}=\frac{O{B}^{{}^{\prime }}}{OB}$
$\therefore$ From equation (ii), $m=\frac{O{B}^{{}^{\prime }}}{OB}\times \frac{D}{u_e}$
or $m=\frac{v_o}{-u_o}\times \frac{D}{-u_e}$ (By sign convention)
where, $u_o\to OB\to$ Distance of object from objective lens, $v_o\to O{B}^{{}^{\prime }}\to$ Distance between objective to the image $A^{{}^{\prime }}{B}^{{}^{\prime }}$
or $m=\frac{-V_{o}D}{u_o{u}_e}$ .......(iii)
When image is formed at least distance of distinct vision:
From formula: $\frac{1}{f}=\frac{1}{v}-\frac{1}{u}$
For eye lens, $\frac{1}{f_e}=\frac{1}{-v_e}-\frac{1}{-u_e}$
or $\frac{1}{f_e}=\frac{1}{-v_e}+\frac{1}{-u_e}$
But, $V_e=D$
$\therefore \frac{1}{f_e}=\frac{1}{D}-\frac{1}{u_e}$
or $\frac{1}{u_e}=\frac{1}{f_e}+\frac{1}{D}$
or $\frac{D}{u_e}=\frac{D}{f_e}+1$
Substituting this value in equation (iii), we get,
$m=\frac{-v_o}{u_o}(1+\frac{D}{f_e})$ ........(iv)
Formula for magnification when image is formed at $D$ magnification.