IPhO 1975 · Theory problem 2

Thick lens

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The focal length \(f\) of a thick glass lens in air with refractive index \(n\), radius curvatures \(r_1\), \(r_2\) and vertex distance \(d\) (see figure) is given by:

\[ f=\frac{nr_1r_2}{(n-1)\left[n(r_2-r_1)+d(n-1)\right]} \]

Figure 1

Remark: \(r_i>0\) means that the central curvature point \(M_i\) is on the right side of the aerial vertex \(S_i\), \(r_i<0\) means that the central curvature point \(M_i\) is on the left side of the aerial vertex \(S_i\) (\(i=1,2\)).

For some special applications it is required, that the focal length is independent from the wavelength.

a) For how many different wavelengths can the same focal length be achieved?

b) Describe a relation between \(r_i\) (\(i=1,2\)), \(d\) and the refractive index \(n\) for which the required wavelength independence can be fulfilled and discuss this relation.

Sketch possible shapes of lenses and mark the central curvature points \(M_1\) and \(M_2\).

c) Prove that for a given planconvex lens a specific focal length can be achieved by only one wavelength.

d) State possible parameters of the thick lens for two further cases in which a certain focal length can be realized for one wavelength only. Take into account the physical and the geometrical circumstances.

Official English text of the problem.

Answers

  • a) At most two distinct wavelengths, assuming that distinct wavelengths have distinct refractive indices in the spectral interval considered.

  • b) For two distinct indices \(n_1,n_2>1\), exact equality of focal lengths requires \[ \boxed{r_1-r_2=d\left(1-\frac{1}{n_1n_2}\right)}. \] First-order wavelength independence at an index \(n_0\) requires \[ \boxed{r_1-r_2=d\left(1-\frac{1}{n_0^2}\right)}. \] Consequently, \(0<r_1-r_2<d\). Possible forms are an unusually thick biconvex lens and two mutually reversed meniscus lenses. Their curvature centres satisfy \(M_1\) to the left of \(M_2\).

  • c) For a planconvex lens with convex-surface radius \(R>0\), \[ \boxed{f=\frac{R}{n-1}}. \] Every attainable focal length therefore determines exactly one refractive index and hence one wavelength.

  • d) Two physically realizable examples, with \(R>0\) an arbitrary length, are \[ \boxed{r_1=R,\quad r_2=-R,\quad d=R} \] and \[ \boxed{r_1=-R,\quad r_2=R,\quad d=R}. \] These are biconvex and biconcave lenses, respectively. For each attainable focal length, the other algebraic index is negative in the first example and less than \(1\) in the second.

Solution

We use the supplied formula within paraxial geometrical optics, with fixed lens geometry and homogeneous glass. Wavelength enters through \(n=n(\lambda)\). For ordinary glass in a transparent interval of normal dispersion, \(n>1\) and \(n(\lambda)\) is one-to-one.

Define the optical power and two convenient constants:

\[ \Phi(n)=\frac1{f(n)},\qquad P=r_1r_2,\qquad C=d+r_2-r_1. \]

For finite, nonzero radii, the given formula becomes

\[ \boxed{ \Phi(n)=\frac{1}{P} \left[Cn-(C+d)+\frac dn\right]. } \tag{1} \]

a) Number of wavelengths

For a prescribed finite focal length \(F\), equation (1) gives

\[ \boxed{ Cn^2-\left(C+d+\frac PF\right)n+d=0. } \tag{2} \]

This is quadratic in \(n\), so there can be at most two distinct refractive indices. If \(C=0\), it becomes linear.

Both roots can be physically admissible: part (b) gives a construction for any chosen pair of distinct glass indices. Thus the maximum is two wavelengths.

Exact constancy throughout a continuous wavelength interval is impossible for a nondegenerate dispersive lens: equation (2) cannot hold identically in \(n\), since its constant term is \(d>0\).

b) Conditions and possible lens shapes

For two distinct indices \(n_1\) and \(n_2\), equality of focal lengths is equivalent to equality of optical powers. Subtracting equation (1) at the two indices,

\[ P\bigl[\Phi(n_1)-\Phi(n_2)\bigr] =(n_1-n_2)\left(C-\frac{d}{n_1n_2}\right). \]

Since \(n_1\ne n_2\),

\[ C=\frac{d}{n_1n_2}. \]

Therefore,

\[ \boxed{ r_1-r_2=d\left(1-\frac{1}{n_1n_2}\right), \qquad d=\frac{n_1n_2}{n_1n_2-1}(r_1-r_2). } \tag{3} \]

Because \(n_1,n_2>1\),

\[ \boxed{0<r_1-r_2<d.} \tag{4} \]

For example, given any two available glass indices, choose \(r_1=R\), \(r_2=-R\), and

\[ d=\frac{2Rn_1n_2}{n_1n_2-1}. \]

This gives a biconvex lens satisfying equation (3), establishing that two physical solutions are possible.

For a fixed lens, the partner index is

\[ n_2=\frac{d}{Cn_1}. \]

Both indices must lie within the actual dispersion range of the glass.

To obtain first-order independence near one wavelength, require

\[ \frac{df}{d\lambda} =\frac{df}{dn}\frac{dn}{d\lambda}=0. \]

Since ordinary dispersion has \(dn/d\lambda\ne0\), this requires \(df/dn=0\), or equivalently \(d\Phi/dn=0\) for finite \(f\). Equation (1) gives

\[ \frac{d\Phi}{dn} =\frac1P\left(C-\frac{d}{n^2}\right). \]

Thus, at the design index \(n_0\),

\[ \boxed{ C=\frac{d}{n_0^2}, \qquad r_1-r_2=d\left(1-\frac1{n_0^2}\right). } \tag{5} \]

The exact two-wavelength condition and the stationary condition are related by

\[ n_0=\sqrt{n_1n_2}. \]

The two distinct indices therefore lie on opposite sides of the stationary index. At the stationary focal length itself, the quadratic has a double root: there is one exact wavelength, with vanishing first-order variation.

The remaining dependence is generally second order, because

\[ \frac{d^2\Phi}{dn^2}=\frac{2d}{Pn^3}\ne0. \]

Substitution into the focal-length formula gives

\[ \boxed{ f(n_0)=-\frac{n_0^2r_1r_2}{d(n_0-1)^2}. } \tag{6} \]

Thus the admissible biconvex form has positive focal length, while the admissible meniscus forms have negative focal length.

For precise descriptions of the sketches, put

\[ x(S_1)=0,\qquad x(S_2)=d. \]

The curvature centres are then

\[ x(M_1)=r_1,\qquad x(M_2)=d+r_2. \]

Under equation (5),

\[ \boxed{x(M_2)-x(M_1)=\frac{d}{n_0^2}>0.} \tag{7} \]

Let \(\delta=d(1-1/n_0^2)\), so \(0<\delta<d\). The following choices give three explicit sketches; the listed points are ordered from left to right.

Lens shape Possible radii Order of vertices and centres
Biconvex: both optical faces bulge outward \(r_1=\delta/2,\quad r_2=-\delta/2\) \(S_1,\ M_1,\ M_2,\ S_2\)
Meniscus: first face convex, second face concave \(r_1=d+\delta,\quad r_2=d\) \(S_1,\ S_2,\ M_1,\ M_2\)
Reversed meniscus: first face concave, second face convex \(r_1=-d,\quad r_2=-(d+\delta)\) \(M_1,\ M_2,\ S_1,\ S_2\)

The optical faces may be small spherical caps joined by a side wall. In the biconvex case, condition (4) becomes

\[ d>r_1+|r_2|. \]

It therefore requires a long piece of glass with convex end faces, rather than the usual thin biconvex shape. A biconcave lens has \(r_1<0<r_2\), contradicting \(r_1>r_2\), so it cannot satisfy the achromatic condition.

c) Planconvex lens

There are two orientations. Taking the plane-surface limit gives

\[ r_2\rightarrow\infty: \qquad f=\frac{r_1}{n-1}, \]

or

\[ r_1\rightarrow\infty: \qquad f=-\frac{r_2}{n-1}. \]

For a convex spherical surface of radius magnitude \(R\), both reduce to

\[ f(n)=\frac{R}{n-1}. \]

The thickness drops out of the effective focal length. Moreover,

\[ \frac{df}{dn}=-\frac{R}{(n-1)^2}<0 \qquad(n>1). \]

Hence \(f(n)\) is strictly monotonic, and a specified attainable focal length determines

\[ \boxed{n=1+\frac RF}. \]

There is therefore exactly one corresponding wavelength within a one-to-one dispersion interval.

d) Two further cases with only one physical wavelength

For \(C\ne0\), equation (2) shows that the product of the two algebraic indices is

\[ n\,\widetilde n=\frac dC. \tag{8} \]

This identifies two useful families.

1. \(r_1-r_2\ge d\), or \(C\le0\).

If \(C<0\), a physical root \(n>1\) has a negative partner,

\[ \widetilde n=\frac{d}{Cn}<0, \]

which cannot describe ordinary glass. If \(C=0\), equation (2) is linear and again has at most one solution.

A concrete example is

\[ r_1=R,\qquad r_2=-R,\qquad d=R. \]

This is a physically realizable biconvex lens. Its focal length is

\[ f(n)=\frac{Rn}{n^2-1}, \]

and

\[ \frac{df}{dn} =-\frac{R(n^2+1)}{(n^2-1)^2}<0. \]

Every attainable positive focal length therefore occurs at one physical index only. Its other algebraic root is \(\widetilde n=-1/n\).

2. \(r_1\le r_2\), or \(C\ge d\).

For a physical root \(n>1\),

\[ 0<\widetilde n=\frac{d}{Cn}\le\frac1n<1. \]

The second root is therefore outside the ordinary-glass range.

A concrete example is

\[ r_1=-R,\qquad r_2=R,\qquad d=R. \]

This is a biconcave lens with positive central thickness; its thickness increases away from the axis. Its focal length is

\[ f(n)=-\frac{Rn}{(n-1)(3n-1)}<0, \]

and its other algebraic index is

\[ \widetilde n=\frac{1}{3n}<1. \]

Thus every attainable negative focal length also corresponds to only one physical wavelength.

Both examples have finite thickness and two curved surfaces. A sufficiently small aperture makes them consistent with the paraxial approximation.

The solution was checked against the official solution.