Optical Instruments: Diffraction

Thursday, 16 June 2011

Compound Microscope and its Magnification

It is an optical device used to observe very small objects like bacteria, cells, gas molecules and section of plants etc.

When ever high magnification is desired, a compound microscope is used. It consists of two convex lenses, one placed before the object called Objective and other placed before the eye called Eye-piece.

The objective has very small focal length $f_{o}$ and small aperture, whereas eye-piece has relatively large focal length $f_{e}$ and large aperture.

Working

When a small object OP is placed at distance p just before the focus of objective then a real, inverted and magnified image IQ is formed at distance q which falls within the focus of eye-piece. The eye-piece is used as magnifying glass and it takes up the image IQ as an object and since it lies within the focal length of eye-piece then it forms a virtual and much magnified final image I'Q' as shown below.

Fig Ray diagram of Compound Microscope

Magnification

If $\alpha$ and $\beta$ are the visual angles subtended by the object and image respectively when formed at the least distance of distinct vision. Then magnifying power of microscope will be

$M=\frac{\beta }{\alpha }$ ${\color{Blue} \Rightarrow (1)}$

where $\alpha=\frac{OP}{d}$ , when object OP is placed at least distance of distinct vision in front of eye. $tan \alpha\approx \alpha$ for small angle.

And in right angle triangle ∆Q'XI'

$tan\beta =\frac{I'Q'}{XQ}$

$\beta =\frac{I'Q'}{d}$

Now putting the values of $\alpha$ and $\beta$ in equation (1) we get,

$\Rightarrow \; M=\frac{I'Q'}{d}\times \frac{d}{OP}$

$\Rightarrow \; M=\frac{I'Q'}{OP}$

It can be written as,

$M=\frac{I'Q'}{IQ}\times \frac{IQ}{OP}$

$M=M_{2}\times M_{1}$ ${\color{Blue} \Rightarrow (2)}$

where

M1 is the magnification of objective

M2 is the magnification of eye-piece

As,

$M_{1}=\frac{q_{o}}{p_{o}}$

And since eye-piece is used as magnifying glass then,

$M_{2}=1+\frac{d}{f_{e}}$

Therefore equation(2) can be written as,

$\Rightarrow \; M=\frac{q_{o}}{p_{o}}\left ( 1+\frac{d}{f_{e}} \right )$ ${\color{Blue} \Rightarrow (3)}$

This expression gives the magnification of eye-piece.
As the object is placed very close to the focus of objective then,

$p_{o}\approx f_{o}$

And since IQ is formed very close to the eye-piece then,

$q_{o}\approx L$

Where L is the distance between objective and eye-piece and it is also known as length of the microscope. Now equation(3) may be written as,

$M\approx \frac{L}{f_{o}}\left ( 1+\frac{d}{f_{e}} \right )$

The limit to which a microscope can be used to resolve details, depends on the width of the objective. A wider objective and use of blue light of short wavelength produces less diffraction and allows more details to be seen.

Resolving Power

The resolving power of an instrument is its ability to reveal the minor details of the object under examination.

When an object is placed in front of a convex lens at a point beyond its focus, a real and inverted image of the object is formed as shown in the figure below:

The ratio of the size of the image to the size of the object is called magnification.

As the object is brought from a far off point to the focus, the magnification goes on increasing. The apparent size of the an object depends on the angle subtended by it at the eyes. Thus, the closer the object is to the eye, the greater is the angle subtended and larger appears the size of the object. The maximum size of an object as seen by naked eye is obtained when the object is placed at the least distance of distinct vision. For lesser distance, the image formed looks blurred and the details of the object are not visible.

The magnifying power or angular magnification is defined as the ratio of the angles subtended by the image as seen through the optical device to that subtended by the object at the unaided eye.

The optical resolution of a microscope or a telescope tells us how close together the two point sources of light can be so that they are still seen as two separate sources. If two point sources are too close, they will appear as one because the optical instrument makes a points source look like a small disc or spot of light with circular diffraction fringes.

Although the magnification can be made as large as one desires by choosing appropriate focal lengths, but the magnification alone is of no use unless we can see the details of the object distinctly.

The resolving power of an instrument is its ability to reveal the minor details of the object under examination.

Resolving Power is expressed as the minimum angle which two point sources subtended at the instrument so that their images are seen two distinct spots of light rather than one. Raleigh showed that for light of wavelength λ through a lens of diameter D, the resolving power is given by

$\alpha _{min}=1.22\frac{\lambda }{D}$

In the case of a grating spectrometer, the resolving power R of the grating is defined as

$R=\frac{\lambda }{\lambda _{2}-\lambda _{1}}=\frac{\lambda }{\Delta \lambda }$

Where λ ≈ λ1 ≈ λ2 and ∆λ = λ2 - λ1.Thus, we see that a grating with high resolving power can distinguish small difference in wavelength. If N is the number of rulings on the grating, it can be shown that the resolving power in the mth-order diffraction equals to the product N x m, i.e.

R = N x m