Question

The magnifying power of a compound microscope can be increased by -

$\left(i\right)$ Using both the lenses of larger focal lengths.

$\left(ii\right)$ Using both the lenses of smaller focal lengths.

$\left(iii\right)$ Decreasing the object distance from the objective lens, but keeping $u_o>f_o$.

• A. Only $\left(i\right)$
• B. $\left(i\right)$ and $\left(ii\right)$
• C. $\left(i\right)$ and $\left(iii\right)$
• D. $\left(ii\right)$ and $\left(iii\right)$

Answer 1

The correct option is D $\left(ii\right)$ and $\left(iii\right)$

We know that,

$\left(a\right)$ Magnification of the compound microscope when the final image is formed at the least distance of distinct vision is given by,

$m=\frac{v_o}{|u_o|}(1+\frac{D}{f_e})$

$\left(b\right)$ Magnification of the compound microscope when the final image is formed at the infinity is given by,

$m=\frac{v_o}{|u_o|}\left(\frac{D}{f_e}\right)$

So, from $\left(a\right)$ and $\left(b\right)$, we can see that, magnification depends on $\frac{1}{f_e}$.

Also, from lens formula, $\frac{1-\frac{v_o}{u_o}}{v_o}=\frac{1}{f_o}$

So, $\frac{v_o}{u_o}$ depends on $\frac{1}{f_o}$.

Thus, again from $\left(a\right)$ and $\left(b\right)$, magnification depends on $\frac{1}{f_o}$.

Therefore, the magnifying power of a compound microscope can be increased by using both the lenses of smaller focal lengths.

Also, from $\left(a\right)$ and $\left(b\right)$, it can be increased by decreasing the object distance from the objective lens $\left(u_o\right)$, but keeping $u_o>f_o$.

Hence, option $\left(D\right)$ is the correct answer.