An Elementary Treatise on Optics/Chapter 9

by Henry Coddington

CHAP. IX.

ABERRATION IN REFRACTION AT SPHERICAL SURFACES.

97. The question here is precisely similar to those we have met before, namely, to determine the difference between the ultimate value of the focal distance for refracted rays, and the value it has for a ray inclined at a sensible though small angle to the axis.

To begin with a single surface. Let v, (Fig. 95.) be the intersection of the refracted ray and the axis, every thing else as before.

Let Av=∆‵.

Referring to the beginning of last Chapter, we find that in strictness

m=QE/QR·Rv/Ev, or QE·Rv=m·RQ·Ev.

Now QR2=	EQ2+ER2+2EQ·ERcosAER
=	(∆−r)2+r2+2(∆−r)r·cosθ
=	∆2−2∆r+2r2+2∆rcosθ−2r2cosθ
=	∆2−2r(∆−r)versinθ.

Similarly, Rv²=∆‵²−2r(∆‵−r)versinθ.

Then putting v for versinθ,

∆−r√∆‵−2r(∆‵−r)v

(1).

=m(∆‵−r)√∆²−2r(∆−r)v

Then proceeding as in Chap. iii., we have

∆‵=∆′+(d∆‵/dv)·v.

To obtain the value d∆‵/dv, we must differentiate the equation (1), which gives

(∆−r)·(∆‵−rv)d∆‵−r(∆‵−r)dv/√∆‵²−2r(∆‵−r)v

=m√∆²−2r(∆−r)v¯¯¯¯¯¯¯¯¯¯·d∆‵−m(∆‵−r)·r(∆−r)dv/√∆²−2r(∆−r)v.

then making v=0, ∆‵=∆′,

(∆−r)·∆′d∆‵−r(∆′−r)dv/∆′

=m∆d∆‵−m(∆′−r)·(∆−r)rdv/∆;

that is, (∆−r){d∆‵−r(∆′−r)/∆‵dv

m∆d∆‵−mr(∆′−r)(∆−r)/∆dv,

or {(m−1)∆+r}d∆‵=(m/∆−1/∆′)r·(∆−r)(∆′−r)dv;

∴ (d∆‵/dv)=r·(∆−r)(∆′−r)/m−1‾‾‾‾‾∆+r(m/∆−1/∆′)

=(∆′−r)²·(m/∆−1/∆′), for ∆′−r=(∆−r)r/m−1‾‾‾‾‾∆+r.

The aberration is therefore (∆′−r)²(m/∆−1/∆′)v.

When the incident rays are parallel, or 1/∆=0, this reduces to

−(∆′−r)²/∆′v, that is, −(F−r)²/Fv.

98. In general, the aberration is positive or negative, that is, Av is greater or less than the ultimate value, according as m/∆−1/∆‵ is positive or negative.

Now m/∆−1/∆′	=m/∆−1/m∆−m−1/mr
	=(m−1/m)1/∆−m−1/mr
	=m2−1/m∆−m−1/mr
	=m−1/m(m+1/∆−1/r).

This is positive, if ∆ be less than (m+1)r, and negative when A is above that value; when ∆=(m+1)r, there is no aberration. See p. 54. Note.

When r is negative, or the surface convex, the aberration is always positive.

99. We will now pass on to the aberration in a lens, (Fig. 96.)

We may consider this as consisting of two parts:

1. The variation in the second focal distance arising from the aberration in the first (α).
2. The additional aberration in the refraction at the second surface (β).

As to the first, we may consider the ratio of the variations as the same with that of the differentials of ∆′ and ∆″.

Now 1/∆″=m/∆′−m−1/r′; ∴ d∆″/d∆′=m∆″²/∆′².

Then since the first aberration is (∆′−r)²(m/∆−1/∆′)·v,

α=m∆″²/∆′²(∆′−r)²(m/∆−1/∆′)·v.

For the second part we must alter our formula by putting

Now 1/m for m, v′ for v, ∆′ for ∆, ∆″ for ∆′, r′ for v.

β=(∆″−r′)²·(1/m∆′−1/∆″)v′.

The whole aberration is therefore

m∆″²/∆′²(∆′−r)²(m/∆−1/∆′)v+(∆″−r′)²·(1/m∆′−1/∆″)v′.

The angles AER, Aer being very nearly the same, we may, without much error, establish that for a particular value of ∆ the aberration varies as v the versed sine of AER, that is, as the square of AR, the radius of the aperture.

Let us examine what kinds of value the aberration in a lens assumes in different cases.

1. For the meniscus or concavo-convex lens we have (r, r′, being both positive.)

The aberration

(A)={m·∆″²/∆′²·(∆′−r)²(m/∆−1/∆′)+(∆″−r′)²(1/m∆′−1/∆″)}v.

Now suppose m=3/2, r=1, r′=5/3, ∆=∞ and therefiore

∆′=	3r=3, ∆″=5.
A=	{−3/2·52/32·22·1/3+102/32(2/3·1/3−1/5)}v
=	−430/81v.

And if v be the versed sine of 2° or .0006, A=−.003, nearly.

Note that the aberration is of a contrary sign to the focal distance, and therefore diminishes it.

2. For the double-concave lens r′ is negative.

A={m·∆″²/∆′²·(∆′−r)²·(m/∆−1/∆′)+(∆″−r′)²·(1/m∆′−1/∆″)}v.

And if

m=	3/2, r=r′=5, ∆=∞, v=.0006,
∆′=	15, ∆″=r=5,
A=	{−3/2·52/152·100/15+100(2/3.15−1/5)}v
=	−{50/45+340/9}v
=	−350/9v
=	−.023

In this case also the aberration diminishes the focal length.

3. For the double-convex lens r is negative,

A={m·∆″²/∆′²·(∆′−r)²·(m/∆−1/∆′)+(∆″−r′)²·(1/m∆′−1/∆″)}·v.

And if

m=	3/2, r=r′=5, ∆=∞, v=.0006,
∆′=	−15, ∆″=−r=−5,
A=	{3/2·52/152·100/15+100(−2/3.15+1/5)}v
=	{10/9+140/9}v
=	150/9v
=	.01

Here the focal distance is negative, and the aberration positive, so that they are still contrary. Observe that the aberration is more than twice as great in the concave lens than in the convex with equal radii.

100.Some writers treat of another aberration arising from that which we have been investigating: it is the distance $qz,$ (Fig. 97.) $Aq$ being the ultimate focal distance and $qz$ perpendicular to $Aq.$

This distance $qz$ is called the lateral aberration, $qo$ the longitudinal,

$qz=qv.{\frac {aS}{av}}=qv{\frac {AR}{Av}},$ nearly.

Since $qv$ varies as the square of $AR,$ it appears that $qz$ varies as its cube.

101.It is important, particularly when a lens is used as a burning-glass to determine whereabouts all the refracted rays are collected within the least space, that is, technically speaking, to find the least circle of aberration or diffusion.

Let $ST$ (Fig. 98.) be the extreme refracted ray on one side: sv^{[errata 1]} a ray on the other side intersecting with this in $n,\ nm$ perpendicular to the axis. Now it is plain that at the maximum state of $mn$, if it has one, all the refracted rays on the same same side with $Sv$ will pass through it, and passing from the section to the actual lens, the circle having $mn$ for its radius will just contain all the rays, so that it will be the circle we seek.

In order to find $mn$ we must know the extreme aberration $qT:$ let this be called $a.$

Let AR	=K which must be measured,
Ar	=k (unknown, or variable,)
Tm	=x,
AT	=T, Av=t.

Since the aberration (longitudinal) varies as the square of the radius of the aperture,

$qv=a.{\frac {k^{2}}{K^{2}}};\quad \therefore vT=a.{\frac {K^{2}-k^{2}}{K^{2}}}..........(1),$

$mn=Tm.{\frac {AR}{AT}}=x.{\frac {K}{T}};\quad \therefore mn\propto x,$ and they are at a maximum together.

vm=mn·vA/Ar=x·K/T·t/k=K/k·t/T·x=K/k·x, nearly. (AT and At are very nearly in a ratio of equality);

(2)

∴ vT=vm+mT=K/kx−x=K+k/Kx

Comparing this with the former value of vT, we find

K+k/Kx=a·K²−k²/K²;

∴ x=ka·K−k/K²=a/K²·k(K−k).

Hence x is at a maximum when k=1/2K; then x=a/4, mn=x·K/T=AR/AT·qT/4.

Since mT=1/4qT, mn=1/4qz; therefore the diameter of the least circle of aberration is equal to half the lateral aberration of the extreme ray.

Its distance from the focus q is three-fourths of the extreme aberration.

Note. What has been proved here for a lens is equally applicable to the cases of reflexion, and refraction at a single surface, as in both of these, the aberration of a ray inclined to the axis varies as the square of the distance of the point of reflexion or refraction from the axis.

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Errata

1. Original: Sv was amended to sv: detail