Question

The focal length of the objective and eye piece of a compound microscope are $1cm$ and $5cm$ respectively. An object is placed at a distance of $1.1cm$ from objective. If final image is formed by at least distance vision, the magnifying power is:

• A. $60$
• B. $50$
• C. $40$
• D. $Noneofthese$

Answer 1

Answer: (B)

Given that,

Focal length of objective $f_o=1cm$

Focal length of eyepiece $f_e=5cm$

Distance of the object $u_o=1.1cm$

Distinct vision = $D$

Now,

The magnifying power of a compound microscope is the product of the linear magnification of the objective and the magnifying power of the eyepiece.

$M.P=M_o\times M_e$

$M.P=\left(\frac{v_0}{v_e}\right)\times \left(\frac{D}{u_0}\right).....\left(I\right)$

Now, for the objective lens

${\frac{1}{f}}_o=\frac{1}{v_o}-\frac{1}{u_o}$

$v_o=\frac{u_o{f}_o}{u_o+f_o}$

$\frac{v_o}{u_o}=\frac{f_o}{u_o+f_o}$

Now, put the value in equation (I)

$M.P=\left(\frac{f_o}{u_o+f_o}\right)\times \left(\frac{D}{u_o}\right)$

Now, because $u_o>f_o$

So,

$M.P=-\left(\frac{f_o}{u_o+f_o}\right)\times \left(\frac{D}{u_o}\right)$

Now, the distance of distinct vision, D may be taken as 25 cm.

So,

$M.P=-\left(\frac{f_o}{u_o+f_o}\right)\times \left(\frac{D}{u_o}\right)$

$M.P=-\left(\frac{1}{1.1+1}\right)\times \frac{25}{5}$

$M.P=-50$

The negative sign indicates that the image is inverted

Hence, the magnifying power is $50$