In a microscope the objective is having focal length $$f_{0}=2 \text{cm}$$ and eye-piece is having focal length $$f_{e} = 4 \text{cm}$$ The tube length is 32 cm. the magnification produced by this microscope for normal adjustment is _______.

For a compound microscope in normal adjustment (where the final image is formed at infinity), the magnifying power ($$M$$) is calculated as:
$$ M = m_0 \times m_e $$

The optical tube length ($$L$$) is considered to be the distance between the focal points, leading to the formula:
$$ M = \frac{L}{f_0} \times \frac{D}{f_e} $$

$$D$$ is the closest distance at which a normal human eye can look at an object clearly without feeling eye strain which is 25cm (universally accepted)
Substituting the given values:
$$ M = \frac{32}{2} \times \frac{25}{4} $$
$$ M = 16 \times 6.25 $$
$$ M = \mathbf{100} $$