The length of the tube of a compound microscope is 15 cm. The focal length of objective and eye lenses are 1 cm and 5 cm respectively. What is the magnifying power of the microscope for the relaxed vision?

The formula is used to calculate length of tube, the focal length of the lens, and the magnification of the emulsion microscope for comfortable viewing through the eyepiece.

The formula used is $m=\frac{L}{f_0}\times \frac{D}{f_e}$

Given the length of the given emulsion microscope,$L=15cm$

Distance of furniture of objective attributes, $f_0=1cm$

The length of focus of eye lenses,$f_e=5cm$

As we know, there is a formula for the improved appearance of the emulsion microscope.

Microscopy,

 $m=\frac{L}{f_0}\times \frac{D}{f_e}$---------1

 Where L is emulsion microscope and $f_0$ indicates the focus length of the object objective.$f_e$ represents the length of the eye focus. Seeing clearly to see the eyes clear $(D=25cm)$

$L=15cm$,$f_0=1cm$,$f_e=5cm$,and $(D=25cm)$ comparison

$m=\frac{15}{1}\times \frac{25}{5}$

$$\begin{gathered} =15\times 5 \\ =75cm \end{gathered}$$

Thus, the increased power of the emulsion microscope for visibility 75.

Hence, the correct option is A.