Question

Describe the construction of a compound microscope. Derive an expression for its total magnification. Draw a ray diagram for the formation of image by a compound microscope.

Answer 1

Construction of Compound microscope:

A schematic diagram of a compound microscope is shown in Fig. The lens nearest the object, called the objective, forms a real, inverted, magnified image of the object. This serves as the object for the second lens, the eyepiece, which functions essentially like a simple microscope or magnifier, produces the final image, which is enlarged and virtual. The first inverted image is thus near (at or within) the focal plane of the eyepiece, at a distance appropriate for final image formation at infinity, or a little closer for image formation at the near point. Clearly, the final image is inverted with respect to the original object.

For magnification, $m_0=\frac{h^{\prime }}{h}=\frac{L}{f_0}$

where we have used the result ,

$tan\beta =\frac{h}{f_0}=\frac{h^{\prime }}{L}$

Here $h\text{'}$ is the size of the first image, the object size being $h$ and fo being the focal length of the objective. The first image is formed near the focal point of the eyepiece. As the first inverted image is near the focal point of the eyepiece.

$m_e=1+\frac{D}{f_e}$

When the final image is formed at infinity, the angular magnification due to the eyepiece.

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

Thus, the total magnification when the image is formed at infinity, is

$m=m_0{m}_e=\frac{L}{f_0}\frac{D}{f_e}$