Focal length of the objective lens, ${\mathrm{f}}_1$=2.0 cm

The focal length of the eyepiece, ${\mathrm{f}}_2$=6.25 cm

(a)
Hint: Use the lens formula.

Step 1: Find the image distance for the objective lens.
Least distance of distinct vision, d’ = 25 cm

$\therefore$ Image distance for the eyepiece, ${\mathrm{v}}_2$​= -25cm

According to the lens formula,

$$$\frac{1}{v_2}-\frac{1}{u_2}=\frac{1}{f_2}$
$\frac{1}{u_2}=\frac{1}{v_2}-\frac{1}{f_2}$
$=\frac{1}{-25}-\frac{1}{6.25}=\frac{-1-4}{25}=\frac{-5}{25}$
$\therefore u_2=-5 cm$

Image distance for the objective lens, $v1=d+u2=15-5=10cm$

Step 2: Find the magnitude of object distance.
According to the lens formula,

$\begin{array}{l}\frac{1}{v_1}-\frac{1}{u_1}=\frac{1}{f_1}\\ \frac{1}{u_1}=\frac{1}{v_1}-\frac{1}{f_1}\\ =\frac{1}{10}-\frac{1}{2}=\frac{1-5}{10}=\frac{-4}{10}\\ \therefore u_1=-2.5 cm\end{array}$

Magnitude of the object distance, $\left|u_1\right|=2.5 cm$

Step 3: Find the magnifying power.
The magnifying power of a compound microscope:

$m=\frac{v_1}{\left|u_1\right|}\left(1+\frac{d\text{'}}{f_2}\right)$
$=\frac{10}{2.5}\left(1+\frac{25}{6.25}\right)=4\left(1+4\right)=20$

Hence, the magnifying power of the microscope is 20.

(b)
Hint: Use the lens formula.

Step 1: Find the image distance for the objective lens.
When the final image is formed at infinity:

$\therefore$ Image distance for the eyepiece, ${\mathrm{v}}_2$​=∞

According to the lens formula,

$$$\begin{array}{l}\frac{1}{v_2}-\frac{1}{u_2}=\frac{1}{f_2}\\ \frac{1}{\infty }-\frac{1}{u_2}=\frac{1}{6.25}\\ \therefore u_2=-6.25 cm\end{array}$

Image distance for the objective lens, $v1=d+u2=15-6.25=8.75cm$

Step 2: Find the magnitude of object distance.
According to the lens formula,

$\begin{array}{l}\frac{1}{v_1}-\frac{1}{u_1}=\frac{1}{f_1}\\ \frac{1}{u_1}=\frac{1}{v_1}-\frac{1}{f_1}\end{array}$
$=\frac{1}{8.75}-\frac{1}{2.0}=\frac{2-8.75}{17.5}$
$\therefore u_1=-\frac{17.5}{6.75}=-2.59 cm$

The magnitude of the object distance, ∣${\mathrm{u}}_1$​∣=2.59 cm

Step 3: Find the magnifying power.
The magnifying power of a compound microscope:

$\begin{array}{l}m=\frac{v_1}{\left|u_1\right|}\left(\frac{d\text{'}}{\left|u_2\right|}\right)\\ =\frac{8.75}{2.59}\times \frac{25}{6.25}=13.51\end{array}$

Hence,  the magnifying power of the microscope is 13.51.