CBSE Class 12-science Answered

Derivation of compound microscope

Asked by phenomenal1ajisback | 20 Feb, 2020, 08:08: PM

Expert Answer

A schematic diagram of a compound microscope is shown in Figure. 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. We now obtain the magnification due to a compound microscope.

The ray diagram of figure shows that the (linear) magnification due to the objective, namely h′/h, equals

begin mathsize 14px style m subscript o space equals space fraction numerator h apostrophe over denominator h end fraction equals L over f subscript o end style

................(1)

where we have used the result , tanβ = h/f_o = h'/L

Here h′ 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. The distance L, i.e., the distance between the
second focal point of the objective and the first focal point of the eyepiece (focal length fe ) is called the tube length

of the compound microscope.

As the first inverted image is near the focal point of the eyepiece, we use the simple microscope magnification equation

to obtain the (angular) magnification m_e .

when the final image is formed at the near point, we have,