It is not the intention of this chapter to study lens aberrations. However, the design of an achromatic doublet lens lends itself to the sort of calculation we are doing in this chapter.

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A combination of two lenses in contact, a converging lens made of crown glass and a weaker diverging lens made of flint glass, can be designed so that the combination is a converging lens that is almost achromatic. Flint glass is a little denser than crown glass, and has a higher refractive index and a greater dispersive power.

The dispersive power \(\omega\) of glass is usually defined as

\[\omega= \frac{n^{(F)}-n^{(C)}}{n^{(D)}-1}.\label{eq:2.10.1} \]

Here C, D and F refer to the wavelengths of the C, D and F Fraunhofer lines in the solar spectrum, which are respectively, H\(\alpha\) (656.3 nm), Na I (589.3 nm), H\(\beta\) (486.1 nm), and which may be loosely referred to as &#;red&#;, &#;yellow&#; and &#;blue&#;. A typical value for a crown glass would be about 0.016, and a typical value for a flint glass would be about 0.028.

An achromatic doublet is typically made of a positive crown glass lens whose power is positive but which decreases with increasing wavelength (i.e. toward the red), cemented to a weaker flint glass lens whose power is negative and also decreases (in magnitude) with increasing wavelength. The sum of the two powers is positive, and varies little with wavelength, going through a shallow minimum. Typically, in designing an achromatic doublet, there will be two requirements to be satisfied: 1. The power or focal length in yellow will be specified, and 2. You would like the power in red to be the same as the power in blue, and to vary little in between.

Consider the doublet illustrated in Figure II.15, constructed of a biconvex crown lens and a biconcave flint lens.

I have indicated the indices and the radii of curvature. The power (reciprocal of the focal length) of the first lens by itself is

\[ P_1 = (n_1-1) \left(\frac{1}{a}+\frac{1}{b}\right),\label{eq:2.10.2} \]

and the power of the second lens is

\[ P_2 = -(n_2-1)\left( \frac{1}{b}+\frac{1}{c}\right). \label{eq:2.10.3} \]

I shall write these for short, in obvious notation, as

\[ P_1 = k_1(n_1-1), \qquad P_2 = -k_2(n_2-1). \label{eq:2.10.4a,b} \]

But we need equations like these for each of the three wavelengths, thus:

\[ P_1^{(C)} = k_1(n_1^{(C)}-1), \qquad P_2^{(C)} = -k_2(n_2^{(C)}-1), \label{eq:2.10.5a,b} \]

\[ P_1^{(D)} = k_1(n_1^{(D)}-1), \qquad P_2^{(D)} = -k_2(n_2^{(D)}-1),\label{eq:2.10.6a,b} \]

\[P_1^{(F)} = k_1(n_1^{(F)}-1), \qquad P_2^{(F)} = -k_2(n_2^{(F)}-1). \label{eq:2.10.7a,b} \]

Now we want to satisfy two conditions. One is that the total power be specified:

\[P_1^{(D)} +P_2^{(D)} = P^{(D)} . \label{eq:2.10.8} \]

The other is that the total power in the red is to equal the total power in the blue, and I now make use of equations \(\ref{eq:2.10.5a,b}\) and \(\ref{eq:2.10.7a,b}\):

\[ k_1(n^{(C)}_1- 1) - k_2(n^{(C)}_2-1)= k_1(n^{(F)}_1- 1) - k_2(n^{(F)}_2-1).\label{eq:2.10.9} \]

On rearrangement, this becomes

\[ k_1(n^{(F)}_1- n^{(C)}_1) = k_2(n^{(F)}_2-n^{(C)}_2). \label{eq:2.10.10} \]

Now, making use of equations \(\ref{eq:2.10.1}\) and \(\ref{eq:2.10.6a,b}\), we obtain the condition that the powers will be the same in red and blue:

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\[ \omega_1P_1 + \omega_2 P_2 = 0. \label{eq:2.10.11} \]

For example, suppose that we want the focal length in yellow to be 16 cm ( \(P^{(D)}= 0.\) cm-1) and that the dispersive powers are 0.016 and 0.028. Equations \(\ref{eq:2.10.8}\) and \(\ref{eq:2.10.11}\) then tell us that we must have \( P_1^{(D)}= 0.\) cm-1 \) and \(P_2^{(D)}= -0.083\) cm-1. (\(f_1 =6.86\) cm and \(f_2 = -12.0\) cm).

If we want to make the first lens equibiconvex, so that \(a = b\), and if \(n_1 = 1.5\), Equation \(\ref{eq:2.10.2}\) tells us that \(a\) = 6.86 cm. If \(n_2 = 1.6\), Equation \(\ref{eq:2.10.3}\) then tells us that \(c = &#;144\) cm. That \(c\) is negative tells us that our assumption that the flint lens was concave to the right was wrong; it is convex to the right.

Exercise \(\PageIndex{1}\)

Suppose that, instead of making the crown lens equibiconvex, you elect to make the last surface flat &#; i.e. \(c\) = &#;. What, then, must \(a\) and \(b\) be?

Answers. \(a\) = 6.55 cm, \(b\) = 7.20 cm.

Achromatic prisms and lenses

Dispersion

When light passes through a prism the amount of deviation depends on the refractive index, and since the refractive index is different for different wavelengths the deviation differs for different colours of light.

If a beam of white light is shone on a prism as shown in Figure 1 the refracted beam is separated into a spectrum (for the present we will restrict ourselves to a consideration of the visible spectrum).

This spreading of the beam is called dispersion and can be shown to depend both on the refracting angle of the prism and on the refractive index of the material of which it is made.
If nR and nB are the refractive indices for red and blue light at the extreme ends of the visible spectrum, then the deviations for red and blue light are:

dR = (nR - 1)A and dB = (nB - 1)A respectively.

Therefore for a prism of small angle the angular dispersion (φ) is given by the formula:

Dispersion (φ) = dR - dB = (nB - nR)A

The mean deviation for a prism is taken as being that produced with yellow light and is given by:

Mean deviation (dY) = (nY - 1)A

where nY is the refractive index of the glass of the prism for yellow light.

'Blue', 'red' and 'yellow' are rather vague terms, however, since each colour represents a range of wavelengths and so for accurate work we choose one particular wavelength within each area of the spectrum:

for red, the C line of hydrogen with a wavelength of 656 nm
for yellow, the D line of sodium with a wavelength of 589 nm
for blue, the F line of hydrogen with a wavelength of 486 nm

The refractive indices of two types of glass for these three standard wavelengths are given in the table below:

  nC nD nF Crown glass 1. 1. 1. Flint glass 1. 1. 1.
The accurate definition for mean deviation therefore becomes:

Mean deviation (dD) = (nD - 1)A

Example problem
Calculate the angular dispersion produced by a flint glass prism of refracting angle 20o. (Take the refractive indices for red and blue right to be as shown in the table above.)

Angular dispersion = (1. - 1.) x 20 = 0.428o

Dispersive power

The mean deviation for a prism is taken as being that produced with yellow light and is given by:where nis the refractive index of the glass of the prism for yellow light.'Blue', 'red' and 'yellow' are rather vague terms, however, since each colour represents a range of wavelengths and so for accurate work we choose one particular wavelength within each area of the spectrum:for red, the C line of hydrogen with a wavelength of 656 nmfor yellow, the D line of sodium with a wavelength of 589 nmfor blue, the F line of hydrogen with a wavelength of 486 nmThe refractive indices of two types of glass for these three standard wavelengths are given in the table below:The accurate definition for mean deviation therefore becomes:

A useful property to consider when calculating the dispersion is the dispersive power of a material. This depends only on the type of material of which a prism or lens is made and not on its shape. Dispersive power is defined as:

Dispersive power (ω) = angular dispersion/mean deviation = (nF &#; nC)/(nD - 1)

Achromatic prisms and lenses

Although the dispersion of white light is useful when we want to look at the spectrum of the light it is a real problem in optical instruments such as telescopes. The lenses in these instruments disperse different colours by different amounts and so bring the different colours to different foci. The images formed are coloured and blurred. It is therefore necessary to deviate the light without dispersing it, and prisms and lenses that do this are called achromatic (Greek, 'without colour').

(a) The achromatic prism
Such a prism is a compound prism made of two prisms of materials with different refractive indices, say n and n'.
The dispersion for prism 1 will be: dR - dB = (nB - nR)A
and that for prism 2: d'R - d'B = (n'B - n'R)A'.

For there to be zero dispersion the algebraic sum of these two dispersions must be zero, and therefore:
(dR - dB) + (d'R - d'B) = (nB - nR)A + (n'B - n'R)A' = 0

Therefore: A/A'= -[n'B - n'R]/[nB - nR]

The negative sign indicates that the prisms must be placed as shown in Figure 2

A single ray of white light passing through an achromatic prism will give rise to a parallel beam of light which when brought to a focus will appear white again. If we take more than one incident ray then the colours will overlap, giving a white centre with coloured edges.

Example problem
A crown glass prism of refracting angle 6o is combined with a flint glass prism to give an achromatic combination.

Calculate the refracting angle of the flint glass prism. What deviation will the compound prism produce? (Take the refractive indices to be those in the table above.)

Let A be the angle of the flint glass prism. Then:
A/6 = - [ 1.523 - 1.515]/[1.665 - 1.643] giving A = - 2.2o

Deviation of red light = (1.515 - 1) x 6 - (1.643 - 1) x 2.2 = 1.68o.

(b) The achromatic lens

For there to be zero dispersion the algebraic sum of these two dispersions must be zero, and therefore:(d- d) + (d'- d') = (n- n)A + (n'- n')A' = 0Therefore: A/A'= -[n'- n']/[n- nThe negative sign indicates that the prisms must be placed as shown in Figure 2A single ray of white light passing through an achromatic prism will give rise to a parallel beam of light which when brought to a focus will appear white again. If we take more than one incident ray then the colours will overlap, giving a white centre with coloured edges.

The dispersion of lenses can be a serious problem in large astronomical instruments - for example, the difference in focal length for red and blue light for a telescope with a mean focal length of around 15 m can be as much as 45 cm. (An exaggerated version of the defect is shown in Figure 3). Such a difference is obviously quite unacceptable when a clearly focused image is required.

This defect of lenses is known as chromatic aberration.

For a lens to be achromatic the focal length for red light (FR) must be the same as that for blue light (FB). As with the achromatic prism this can be produced by using a 'doublet' made of two thin lenses of different refractive indices (Figure 4).

For blue light: 1/FB = 1/fB + 1/fB' For red light: 1/FR = 1/fR + 1/fR'

and also we have for each lens:

1/fB - 1/fR = (nB &#; nR)(1/R1 + 1/R2) and 1/fy = (nY &#; 1)(1/R1 + 1/R2)

Therefore:

1/fB - 1/fR = w/fY and 1/fB' &#; 1/fR' = w'/fY' This gives: w/fY + w'/fY' = 0


Therefore:

ω/fY + ω'/fY' = 0

In this formula the negative sign means that one of the lenses is convex and the other concave.
Notice that we have only made the lens truly achromatic for two colours, red and blue. There will still be a spread of colour due to the other wavelengths.

It is possible to make an achromatic lens using two thin lenses of the same material if they are separated by a distance equal to the mean of their focal lengths.

Defects of lenses

This defect of lenses is known as chromatic aberration.For a lens to be achromatic the focal length for red light (F) must be the same as that for blue light (F). As with the achromatic prism this can be produced by using a 'doublet' made of two thin lenses of different refractive indices (Figure 4).For blue light: 1/F= 1/f+ 1/f' For red light: 1/F= 1/f+ 1/fand also we have for each lens:1/f- 1/f= (n&#; n)(1/R+ 1/R) and 1/f= (n&#; 1)(1/R+ 1/RTherefore:1/f- 1/f/fand 1/f' &#; 1/f' ='/f' This gives:/f'/f' = 0Therefore:In this formula the negative sign means that one of the lenses is convex and the other concave.Notice that we have only made the lens truly achromatic for two colours, red and blue. There will still be a spread of colour due to the other wavelengths.It is possible to make an achromatic lens using two thin lenses of the same material if they are separated by a distance equal to the mean of their focal lengths.

In addition to chromatic aberration described above, lenses suffer from several other defects.
(a) Spherical aberration
This is a result of the inner and outer portions of a lens having different focal lengths, that of the outside being shorter than that of the centre.
One way of reducing this is to make the deviation at the two surfaces as nearly equal as possible. Spherical aberration is therefore particularly marked when using a piano-convex lens with parallel light hitting the plane face.
Spherical aberration is also reduced by decreasing the aperture of a lens and by increasing its focal length.

(b) Coma
This defect produces a comet-like tail added to all images. It results from off-axis objects coupled with the different magnifications of different zones of the lens.
The rays from the vertical plane intersect in a horizontal line while those from a horizontal plane intersect in a vertical line.

(c) Astigmatism
If the object point lies off the axis of the lens then the rays from the horizontal and vertical planes come to a focus at different distances from the lens.

(d) Distortion
The magnification of the lens varies from its centre to its edge and so the magnification of the image will vary as well. This gives rise to distortion.

 

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