Page:EB1911 - Volume 16.djvu/444

refracting surface. Let such a surface divide media of refractive indices n and n′, the former being to the left. The point where the axis intersects the surface is the vertex S (fig. 7). Denote the distance of the axial object-point O from S by s; the distance from O to the point of incidence P by p; the radius of the spherical surface by r; and the distance OC by c, C being the centre of the sphere. Let u be the angle made by the ray with the axis, and i the angle of incidence, i.e. the angle between the ray and the normal to the sphere at the point of incidence. The corresponding quantities in the image-space are denoted by the same letters with a dash. From the triangle O′PC we have sin u = (r/c) sin i, and from the triangle O′PC we have sin u′ = (r/c′) sin i′. By Snell’s law we have n′/n = sin i/sin i′, and also φ = u′ + i′. Consequently c′ and the position of the image may be found.

Fig. 7.

To determine whether all the rays proceeding from O are refracted through O′, we investigate the triangle OPO′. We have p/p′ = sin u′/sin u. Substituting for sin u and sin u′ the values found above, we obtain p′/p = c′ sin i/c sin i′ = n′c′/nc. Also c = OC = CS + SO = −SC + SO = s − r, and similarly c′ = s′ − r. Substituting these values we obtain

p′/p = n′(s′ − r)/n(s − r), or n(s − r)/p = n′(s′ − r)/p .

(6)

To obtain p and p′ we use the triangles OPC and O′PC; we have p² = (s − r)² + r² + 2r(s − r) cos φ, p′² = (s′ − r)² + r² + 2r(s′ − r) cos φ. Hence if s, r, n and n′ be constant, s′ must vary as φ varies. The refracted rays therefore do not reunite in a point, and the deflection is termed the spherical aberration (see Aberration).

Developing cos φ in powers of φ, we obtain

p² = (s − r)² + r ² + 2r(s − r) ${\Big \{}$ 1 − φ²/2! + φ⁴/4! − φ⁶/6! + . . . ${\Big \}}$ ,

and therefore for such values of φ for which the second and higher powers may be neglected, we have p² = (s − r)² + r² + 2r(s − r), i.e. p = s, and similarly p′ = s′. Equation (6) then becomes n(s − r)/s = n′(s′ − r)/s′ or

n′/s′ = n/s + n′ − n/r .

(7)

This relation shows that in a very small central aperture in which the equation p = s holds, all rays proceeding from an object-point are exactly united in an image-point, and therefore the equations previously deduced are valid for this aperture. K. F. Gauss derived the equations for thin pencils in his Dioptrische Untersuchungen (1840) by very elegant methods. More recently the laws relating to systems with finite aperture have been approximately realized, as for example, in well-corrected photographic objectives.

Position of the Cardinal Points of a Lens.—Taking the case of a single spherical refracting surface, and limiting ourselves to the small central aperture, it is seen that the second principal focus F′ is obtained when s is infinitely great. Consequently s′ = -f ′; the difference of sign is obvious, since s′ is measured from S, while f ′ is measured from F′. The focal lengths are directly deducible from equation (7):—

f ′ = n′r / (n′ − n)

(8)

f = nr / (n′ − n).

(9)

By joining this simple refracting system with a similar one, so that the second spherical surface limits the medium of refractive index n′, we derive the spherical lens. Generally the two spherical surfaces enclose a glass lens, and are bounded on the outside by air of refractive index 1.

The deduction of the cardinal points of a spherical glass lens in air from the relations already proved is readily effected if we regard the lens as a combination of two systems each having one refracting surface, the light passing in the first system from air to glass, and in the second from glass to air. If we know the refractive index of the glass n, the radii r₁, r₂ of the spherical surfaces, and the distances of the two lens-vertices (or the thickness of the lens d) we can determine all the properties of the lens. A biconvex lens is shown in fig. 8. Let F₁ be the first principal focus of the first system of radius r₁, and F₁′ the second principal focus; and let S₁ be its vertex. Denote the distance F₁ S₁ (the first principal focal length) by f₁, and the corresponding distance F′₁ S₁ by f ′₁. Let the corresponding quantities in the second system be denoted by the same letters with the suffix 2.

By equations (8) and (9) we have

f1 =	r1	,  f ′1 = −	nr1	,  f2 = −	nr2	,  f ′2 =	r2	,
	n − 1		n − 1		n − 1		n − 1

f₂ having the opposite sign to f₁. Denoting the distance F′₁F₂ by Δ, we have Δ = F′₁F₂ = F′₁S₁ + S₁S₂ + S₂F₂ = F′₁S₁ + S₁S₂ − F₂S₂ = f ′₁ + d − f₂. Substituting for f ′₁ and f₂ we obtain

Δ = nr₁/n − 1 + d + nr₂/n − 1

Writing R = Δ(n − 1), this relation becomes

R = n(r₂ − r₁) + d(n − 1).

We have already shown that f (the first principal focal length of a compound system) = −f₁f₂/Δ. Substituting for f₁, f₂ and Δ the values found above, we obtain

f =	r1r2n	=	r1r2n	,
	(n − 1)R		(n − 1) {n (r2 − r1) + d(n − 1)}

(10)

which is equivalent to

1	= (n − 1) ${\Big \{}$	1	−	1	${\Big \}}$ +	(n − 1)2 d	.
f		r1		r2		r1r2n

If the lens be infinitely thin, i.e. if d be zero, we have for the first principal focal length.

1	= (n − 1) ${\Big \{}$	1	−	1	${\Big \}}$.
f		r1		r2

By the same method we obtain for the second principal focal length

f ′ =	f ′1f ′2	= −	nr1r2	= −f.
	Δ		(n − 1)R

Fig. 8.

The reciprocal of the focal length is termed the power of the lens and is denoted by φ. In formulae involving φ it is customary to denote the reciprocal of the radii by the symbol ρ; we thus have φ = 1/f, ρ = 1/r. Equation (10) thus becomes

φ = (n − 1) (ρ1 − ρ2) +	(n − 1)2 dρ1ρ2	.
	n

The unit of power employed by spectacle-makers is termed the diopter or dioptric (see Spectacles).

We proceed to determine the distances of the focal points from the vertices of the lens, i.e. the distances FS₁ and F′S₂. Since F is represented by the first system in F₂, we have by equation (2)

x1 =	f1f ′1	=	f1f ′1	= −	nr12	,
	x′1		Δ		(n − 1)R

where x₁ = F₁F, and x′₁ = F′₁F₂ = Δ. The distance of the first principal focus from the vertex S, i.e. S₁F, which we denote by s_F is given by s_F = S₁F = S₁F₁ + F₁F = −F₁S₁ + F₁F. Now F₁S₁ is the distance from the vertex of the first principal focus of the first system, i.e. f₁ and F₁F = x₁. Substituting these values, we obtain

sF = −	r1	−	nr12	= −	r1 (nr1 + R)	.
	n − 1		(n − 1)R		(n − 1)R

The distance F′₂F′ or x′₂ is similarly determined by considering F′₁ to be represented by the second system in F′.

We have

x′2 =	f2f ′2	= −	f2f ′2	=	nr22	,
	x2		Δ		(n − 1)R

so that

sF′ = x′2 − f ′2 =	r2 (nr2 − R)	,
	(n − 1)R

where s_F′ denotes the distance of the second principal focus from the vertex S₂.

The two focal lengths and the distances of the foci from the vertices being known, the positions of the remaining cardinal points, i.e. the principal points H and H′, are readily determined. Let s_H = S₁H, i.e. the distance of the object-side principal point from the vertex of the first surface, and s_H′ = S₂H′, i.e. the distance of the image-side principal point from the vertex of the second surface, then f = FH = FS₁ + S₁H = −S₁F + S₁H = −s_F + s_H; hence s_H = s_F + f = −dr₁/R. Similarly s_H′ = s_F′ + f ′ = −dr₂/R. It is readily seen that the distances s_H and s_H′ are in the ratio of the radii r₁ and r₂.

The distance between the two principal planes (the interstitium) is deduced very simply. We have S₁S₂ = S₁H + HH′ + H′S₂, or HH′ = S₁S₂ − S₁H + S₂H′. Substituting, we have

HH′ = d − s_H + s_H′ = d(n − 1) (r₂ − r₁ + d)/R.

The interstitium becomes zero, or the two principal planes coincide, if d = r₁ − r₂.

We have now derived all the properties of the lens in terms of its elements, viz. the refractive index, the radii of the surfaces, and the thickness.

Forms of Lenses.—By varying the signs and relative magnitude of the radii, lenses may be divided into two groups according to their action, and into four groups according to their form.

According to their action, lenses are either collecting, convergent