Compound Microscope:
A compound microscope consists of two lenses: the objective lens and the eyepiece. The objective lens forms a real, inverted, and magnified image of the object, and the eyepiece magnifies this image further.

- The object is placed slightly inside the focal point of the objective lens.
- The objective lens produces a real, inverted, and magnified image, which acts as the object for the eyepiece.
- The eyepiece forms an image at infinity, as the final image is formed at infinity (in a relaxed eye condition).
The total magnification \( M_{\text{total}} \) of the microscope is the product of the magnification produced by the objective lens \( M_{\text{obj}} \) and the magnification produced by the eyepiece \( M_{\text{eyepiece}} \):
\[ M_{\text{total}} = M_{\text{obj}} \times M_{\text{eyepiece}} \]
The magnification of the objective lens is given by:
\[ M_{\text{obj}} = - \frac{v_{\text{obj}}}{u_{\text{obj}}} \]
Where \( v_{\text{obj}} \) is the image distance and \( u_{\text{obj}} \) is the object distance for the objective lens.
The magnification of the eyepiece is given by:
\[ M_{\text{eyepiece}} = \frac{D}{f_{\text{eyepiece}}} \]
Where \( D \) is the near point distance of the eye (usually taken as 25 cm), and \( f_{\text{eyepiece}} \) is the focal length of the eyepiece.
Thus, the total magnification is the product of these two magnifications.

A ladder of fixed length \( h \) is to be placed along the wall such that it is free to move along the height of the wall.
Based upon the above information, answer the following questions:
(iii) (b) If the foot of the ladder, whose length is 5 m, is being pulled towards the wall such that the rate of decrease of distance \( y \) is \( 2 \, \text{m/s} \), then at what rate is the height on the wall \( x \) increasing when the foot of the ladder is 3 m away from the wall?