Name: Dual Nature of Radiation
Uploaded: 2019-07-19
Duration: 3 min 36 s
Description: What is the nature of light? Is it a wave or a particle? The answer is light has a dual nature. Light can be treated as a particle as well as a wave.

Question

What is achromatism? Derive a condition to achieve achromatism for a lens combination. State the conditions for it to be converging.

Accepted Answer

Achromatism is the condition for a combination of two thin lenses to minimise or eliminate the chromatic aberration for two specific colours. An achromatic lens can be defined as a lens which is made by a combination of two different types of lenses carrying different focal powers in a manner such that the images formed by the combined lenses are free from chromatic aberrations and this process is called achromatism.

The most commonly used and the earliest example of an achromatic lens is the achromatic doublet. It is a pair of glasses, of which, one is concave and the other is convex. The concave element of the doublet is composed of flint glass and the convex element is made of crown glass. These are placed next to each other in such a manner that the chromatic aberration of one element is balanced by the chromatic aberration of another.

Condition for achromatism: An achromatic combination aims to eliminate the total dispersion for two specific colours by using materials of different dispersive powers, $ω_{1}$ and $ω_{2}$ , forming the lenses. By the lens-makers formula,

$\frac{1}{f} = (n - 1) = (\frac{1}{R_{1}} - \frac{1}{R_{2}}) = (n - 1) K$

Abbreviating the notation for brevity, for the two lenses,

$- \frac{1}{f_{1}} = (n_{1} - l) K_{1}$ and $\frac{1}{f_{2}} = (n_{2} - l) K_{2}$

For each lens of the combination, $n$ and $f$ vary with colour while $K$ is constant. For two lenses

in contact,

$\frac{1}{f} = \frac{1}{f_{1}} + \frac{1}{f_{2}} = (n_{1} - 1) K_{1} + (n_{2} - 1)$

$K_{2} = (n_{1} K_{1} + n_{2} K_{2}) - (K_{1} + K_{2})$

This expression will give the focal length of the achromatic doublet for red ( $f_{2 R}$ ) and violet ( $f_{2 V}$ ) when the appropriate refractive indices are used. For $f_{R}$ = $f_{V}$ ,

$\frac{1}{f_{1 R}} + \frac{1}{f_{2 R}} = \frac{1}{f_{1 v}} + \frac{1}{f_{2 v}}$

$∴ (n_{1 R} K_{1} + n_{2 R} K_{2}) - (K_{1} + K_{2})$ $= (n_{1 v} K_{1} + n_{2 v} K_{2}) - (K_{1} + K_{2})$

$∴ (n_{1 v} - n_{1 R}) K_{1} = (n_{2 v} - n_{2 R}) K_{2}$

$\frac{K_{1}}{K_{2}} = - \frac{n_{2 v} - n_{2 R}}{n_{1 v} - n_{1 R}}$ ….. ( $1$ )

The focal length of the doublet can conveniently be specified as $f_{Y}$ , the focal length for yellow light, which is roughly midway between the red and violet extremes.

$\frac{1}{f_{1 V}} = (n_{1 Y} - 1) K_{1}$ and $\frac{1}{f_{2 Y}} = (n_{2 Y} - 1) K_{2} $

$∴ \frac{f_{2 Y}}{f_{1 Y}} = - \frac{n_{1 Y} - 1}{n_{2 Y} - 1} . \frac{K_{1}}{K_{2}}$

$∴ \frac{f_{2 Y}}{f_{1 Y}} = - \frac{n_{2 V} - n_{2 R}}{n_{2 Y} - 1} . \frac{n_{1 Y} - 1}{n_{1 V} - n_{1 R}}$ [from Eq.(1)]

$∴ \frac{f_{2 Y}}{f_{1 Y}} = \frac{ω_{2}}{ω_{1}}$ ... ( $3$ )

Equation ( $3$ ) is the condition for achromatism :

Since dispersive powers are always positive, one lens of the achromatic doublet must be a positive or converging lens and the other a negative or diverging lens.

If the second lens is a concave lens, $f_{2 Y}$ is negative.

$∴ \frac{1}{f_{Y}} + \frac{1}{f_{1 Y}} = \frac{1}{f_{2 V}}$

For the combination to be converging, i.e., for

$f_{Y} > 0$ , $f_{1 Y} < f_{2 Y}$ so that, from Eq.( $3$ ),

$ω_{2} > ω_{1}$

Thus, the first lens is a convex lens of lower dispersive power than the second lens which is a concave lens.

What is achromatism? Derive a condition to achieve achromatism for a lens combination. State the conditions for it to be converging.

Important Questions on Optics

Learn and Prepare for any exam you want !