Chromatic Aberrations in Lenses

Chromatic Aberrations in Lenses

The inability of a lens to focus all colours of light at a single point is called chromatic aberration or axial or longitudinal chromatic aberration. It is measured by the difference in focal lengths between red and violet colours.

\begin{align*} \text {Chromatic aberration} &= f_r – f_v \\ \end{align*}

Using lens maker’s formula, for mean colour of light, we have

\begin{align*} \frac 1f &= (\mu – 1) \left ( \frac {1}{R_1} + \frac {1}{R_2} \right ) \\ \text {or,} \: \frac {1}{R_1} + \frac {1}{R_2} &= \frac {1}{f(\mu -1)} \dots (i) \\ \end{align*}

Where f is focal length of mean colour, µ is refractive index of mean colour, R₁ and R₂ are radii of curvature of two lens surfaces.

For violet colour, we have

\begin{align*} \frac 1f &= (\mu _v – 1) \left ( \frac {1}{R_1} + \frac {1}{R_2} \right ) \\ \text {or,} \: \frac {1}{f_v} &= (\mu _v – 1) \frac {1}{f(\mu -1)}\\ \text {or,} \: \frac {1}{f_v} &= \frac {u_v - 1}{f(\mu -1)} \dots (ii) \\ \text {where} \: \mu _v \end{align*}is refractive index of violet colour. Similarly for red colour, we have \begin{align*}\frac {1}{f_r} &=\frac {\mu_v - 1}{f(\mu -1)} \dots (iii)\\ \end{align*}

Here \(\mu _r \)is refractive index of red colour. Subtracting equation (iii)from equation (ii) we get \begin{align*}\frac {1}{f_v} - \frac {1}{f_r} &= \frac {\mu_v - 1}{f(\mu -1)} - \frac {\mu_r - 1}{f(\mu -1)}\\ \text {or,} \: \frac {f_r – f_v}{f_v . f_r} &= \frac { u_v – 1 - \mu _r + 1}{f(\mu -1)} \\ \text {or,} \: f_f – f_v &= \frac {(\mu _v -\mu_r) f_vf_r}{f(\mu – 1)} \dots (iv) \\ \end{align*}

Achromatic Combination of Lenses

The combination of two thin lenses in which their combination is free from chromatic aberration is called the achromatic combination of lenses.

Consider two thin lenses l and L’ of dispersive power \(\omega \text {and} \omega ‘\) respectively placed in contact with each other as shown in the figure. Let \(\mu, \: \mu \text {and}\: \mu _r\) are the refractive indices of L for violet, mean and red colour respectively, and f_v , f and f_r are the focal lengths of respective colours. Similarly \(\mu, \: \mu \text {and}\: \mu _r ; f_v’, f’, f_r’ \) are corresponding quantities of L’.

For lens L, focal length of mean colour is

\begin{align*} \frac 1f &= (\mu – 1) \left ( \frac {1}{R_1} + \frac {1}{R_2} \right ) \\ \text {or,} \: \frac {1}{R_1} + \frac {1}{R_2} &= \frac {1}{f(\mu -1)} \\ \end{align*}

where R₁ and R₂ are radii of curvature of two lens surfaces. Focal length of lens L for violet colour is

\begin{align*} \frac 1f_v &= (\mu _v – 1) \left ( \frac {1}{R_1} + \frac {1}{R_2} \right ) \\\text {or,} \: \frac {1}{f_v} &= \frac {\mu_v - 1}{f(\mu -1)} \dots (i) \\ \end{align*}Similarly, focal length of lens L’ for violet colour\begin{align*} \frac {1}{f_v’} &= \frac {\mu_v’ - 1}{f’(\mu’ -1)} \dots (ii) \\ \text {If} \: F_v \end{align*}is the combined focal length of two lenses for violet colour, then \begin{align*}\frac {1}{F_v} &= \frac {1}{f_v} + \frac {1}{f_v’} \dots (iii) \\ \frac {1}{F_v} &= \frac {\mu_v - 1}{f(\mu -1)} + \frac {\mu_v’ - 1}{f(\mu ‘-1)} \dots (iv) \\ \end{align*}In the same way for red colour,\begin{align*}\frac {1}{F_r} &= \frac {\mu_r - 1}{f(\mu -1)} + \frac {\mu_r’ - 1}{f(\mu ‘-1)} \dots (v)\\ \end{align*}

\begin{align*} \text {For achromatic combination, we have} \\ F_r &= F_v \\ \text {or,} \: \frac {1}{F_v} &= \frac {1}{F_r} \\ \text {or,} \: \frac {\mu_v - 1}{f(\mu -1)} + \frac {\mu_v’ - 1}{f(\mu ‘-1)} &= \frac {\mu_r - 1}{f(\mu -1)} + \frac {\mu_r’ - 1}{f(\mu ‘-1)} \\ \text {or,} \: \frac {\mu_v - 1}{f(\mu -1)} - \frac {\mu_r - 1}{f(\mu -1)} &= \frac {\mu_r’ - 1}{f(\mu’ -1)} - \frac {\mu_v’ - 1}{f(\mu ‘-1)} \\ \text {or,} \: \frac {\mu _v – 1 - \mu _r + 1}{f(\mu – 1)} &= \frac {\mu _v’ – 1 - \mu _r’ + 1}{f(\mu’ – 1)} \\ \text {or,} \: \frac {\mu_v - \mu_r}{f(\mu -1)} &= \frac {\mu_v’ - \mu_r’}{f(\mu ‘-1)} \\ \text {or,} \: \frac {\omega }{f} &= - \frac {\omega ‘}{f} \\ \text {where} \: \frac {\mu_v - \mu_r}{f(\mu -1)} = \omega \: \text {and} \: \frac {\mu_v’ - \mu_r’}{f(\mu ‘-1)} = \omega ‘\\ \therefore \frac {\omega }{f} + \frac {\omega ‘}{f} &= 0 \\ \end{align*}

This is the condition for achromatic combination of two lenses.