**2016:**

The magnifying power of a compound microscope is given by

```
M = m_o \times m_e
```

where $m_o$ is the magnification of the objective lens and $m_e$ is the magnification of the eyepiece. The magnification of the objective lens is given by

```
m_o = \frac{v}{u}
```

where $v$ is the image distance and $u$ is the object distance. The magnification of the eyepiece is given by

```
m_e = \frac{D}{f_e}
```

where $D$ is the distance of distinct vision and $f_e$ is the focal length of the eyepiece.

In this problem, we are given that $f_o = 1.25$ cm, $u = 1.5$ cm, $f_e = 5$ cm, and $D = 25$ cm. We can find $v$ using the thin lens equation:

```
\frac{1}{f} = \frac{1}{v} + \frac{1}{u
```