Name: Total Internal Reflection From Prism
Uploaded: 2019-07-19
Duration: 1 min 56 s
Description: You must have seen reflectors on a bicycle. Why are they attached to the mudguards? What use do they have? Let’s watch this video on reflectors to find out.

Question

Draw a labelled diagram of simple microscope and establish an expression for the magnifying power of simple microscope, when the image formed at

$(i)$ Infinity

Accepted Answer

A simple microscope has a limited maximum magnification (≤ 9) for realistic focal lengths.

For much larger magnifications, one uses two lenses, one compounding the effect of the other.

This is known as a compound microscope.

A schematic diagram of a compound microscope is shown in above 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 formationat infinity, or a little closer for image formationat 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 given in figure shows that the (linear) magnification m_o due to the objective, namely $h^{'} / h$ , equals

$m_{0} = \frac{h^{'}}{h} = \frac{L}{f_{0}}$

To get magnification m_o due to objective as given above, we have used the properties of similar triangles appearing between objective lens and first image as given below.

$tanβ = \frac{h}{f_{0}} = \frac{h^{'}}{L}$

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 $f_{e}$ ) 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 magnification expression for the simple microscope to obtain the (angular) magnification m_e.

When the final image is formed at the near point, is

$m_{e} = (1 + \frac{D}{f_{e}})$

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

$m_{e} = \frac{D}{f_{e}}$

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

$m = m_{o} m_{e} = (\frac{L}{f_{o}}) (\frac{D}{f_{e}})$

Clearly, to achieve a large magnification of a small object (hence the name microscope), the objective and eyepiece should have small focal lengths.

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