Hey, do you want to learn about achromatism in lenses? If yes. Then keep reading

## Achromatism:

We have just that when a white object is placed in front of a lens, then its images of different colors are formed at different positions and are of different sizes. These defects are called ‘longitudinal chromatic aberration’ and ‘lateral chromatic aberration’ respectively. If two or more lenses be different colors are in the same position and of the same size, then the combination is called ‘achromatic combination of lenses, and this property is called ‘achromatism’.

In practice, both types of chromatic aberrations cannot be removed for all colors. We can remove both types of chromatic aberration only for two colors by placing in contact two lenses of appropriate focal lengths and of an appropriate different material. On the other hand, only lateral chromatic aberration can be removed for all colors when two lenses of an appropriate different material. On the other hand, only lateral chromatic aberration can be removed for all colors when two lenses of the same material are placed at a particular distance apart.

## Condition of Achromatism for two thin lenses in contact:

Suppose two thin lenses are placed in contact. Suppose the dispersive powers of the materials of these lenses between violet and red respectively $n_{V}, n_{R}, n_{y}$ and $\mathrm{n}_{\mathrm{V}}^{\prime}, \mathrm{n}_{\mathrm{R}}^{\prime}, \mathrm{n}_{\mathrm{y}}^{\prime}$. If for these rays the focal lengths of the first lens are respectively $f_{v}, f_{R}, f_{y}$ and the focal lengths of the second lens are $f_{V}^{\prime}, f_{R}^{\prime}, f_{y}^{\prime}$, then for the first lens, we have

$\frac{1}{f_{V}}=\left(n_{V}-1\right)\left[\frac{1}{R_{1}}-\frac{1}{R_{2}}\right]$…..(1)

$\frac{1}{f_{R}}=\left(n_{R}-1\right)\left[\frac{1}{R_{1}}-\frac{1}{R_{2}}\right]$…..(2)

Subtracting the second equation from the first, we get

$\frac{1}{f_{V}}-\frac{1}{f_{R}}=\left(n_{V}-n_{R}\right)\left[\frac{1}{R_{1}}-\frac{1}{R_{2}}\right]$

$=\frac{\left(n_{V}-n_{R}\right)}{\left(n_{y}-1\right)}\left(n_{y}-1\right)\left[\frac{1}{R_{1}}-\frac{1}{R_{2}}\right]$

$\frac{1}{f_{v}}-\frac{1}{f_{r}}=\omega \frac{1}{f_{y}}$….(3)

because $\frac{n_{v}-n_{R}}{n_{y}-1}=\omega$

and $\left(n_{y}-1\right)\left(\frac{1}{R_{1}}-\frac{1}{R_{2}}\right)=\frac{1}{f_{y}}$

Similarly, for the second lens, we have

$\frac{1}{f_{v}^{\prime}}-\frac{1}{f_{R}^{\prime}}=\omega \frac{1}{f_{y}^{\prime}}$….(4)

Adding equations (3) and (4), we get

$\left(\frac{1}{f_{V}}+\frac{1}{f_{V}^{\prime}}\right)-\left(\frac{1}{f_{R}}+\frac{1}{f_{R}^{\prime}}\right)=\frac{\omega}{f_{y}}+\frac{\omega^{\prime}}{f_{y}^{\prime}}$……(5)

If the focal lengths of this lens-combination for the violet and the red rays be $\mathrm{F}_{\mathrm{V}}$ and $\mathrm{F}_{\mathrm{R}}$ respectively, then

$\frac{1}{f_{V}}+\frac{1}{f_{V}^{\prime}}=\frac{1}{F_{V}}$

And

$\frac{1}{f_{R}}+\frac{1}{f_{R}^{\prime}}=\frac{1}{F_{R}}$ from eq. (5), we have $\frac{1}{F_{V}}-\frac{1}{F_{R}}=\frac{\omega}{f_{y}}+\frac{\omega}{f_{y}^{\prime}}$

But for the achromatism of the lens combination, the focal length must be the same for all colors of light i.e. $F_{v}=F_{R}$. Hence from the above equation, we have

$\frac{\omega}{f_{y}}+\frac{\omega^{\prime}}{f_{y}^{\prime}}=0$…..(6)

Or

$\frac{\omega}{f_{y}}=-\frac{\omega^{\prime}}{f_{y}^{\prime}}$…..(7) **This is the condition for a lens combination to achromatic. It gives us the following information**

- Both the lenses should be of a different material. If both the lenses are of the same material, then $\omega=\omega^{\prime}$ and then from equation (6). We have

$\frac{1}{f_{y}}+\frac{1}{f_{y}^{\prime}}=0$

or

$\frac{1}{F_{y}}=0$ or $F_{y}=\infty$

that is the combination will then behave like a plane glass-plate. - $\omega$ and $\omega^{\prime}$ are positive quantities. Hence, according to eq. (7), $\mathrm{f}_{\mathrm{y}}$ and $\mathrm{f}_{\mathrm{y}}^{\prime}$ should be of opposite signs, i.e., if one lens is convex, the other should be concave.
- For the combination of behavior like a convergent (convex) lens–system, the power of the convex lens should be greater than that of the concave lens. On other words, the focal length of the convex lens should be smaller than the concave lens. According to eq. (7), we have

$\frac{\mathrm{f}_{\mathrm{y}}}{\mathrm{f}_{\mathrm{y}}^{\prime}}=-\frac{\omega}{\omega^{\prime}}$ If $\mathrm{f}_{\mathrm{y}}$ is less than $\mathrm{f}_{\mathrm{y}}^{\prime}$

then $\omega$ should be less than $\omega^{\prime}$. Hence for a converging lens system, the convex lens should be made of a material of smaller dispersive power.

The dispersive power of crown glass is smaller than that of flint glass. Hence in an achromatic lens-doublet, the convex lens is of crown glass and the concave lens is of flint glass, and they are cemented together by Canada Balsam (a transparent cement). This achromatic combination is used in optical instruments such as microscopes, telescopes, cameras, etc.

In the condition for achromatism

$\frac{\mathrm{f}_{\mathrm{y}}}{\mathrm{f}_{\mathrm{y}}^{\prime}}=-\frac{\omega}{\omega^{\prime}}$

So, that’s all from this article. I hope you get the idea about achromatism in lenses. If you found this article informative then please share it with your friends. If you have any confusion related to this topic, then you can ask in the comments section down below.

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