IPhO 1972 · Theory problem 4

Plano-convex lens between two media; Billet split lens

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A thin lens plane-convex with the diameter \(2r\), the curvature radius \(R\) and the refractive index \(n_o\) is positioned so that on its left side is air (\(n_1=1\)), and on its right side there is a transparent medium with the refractive index \(n_2\ne 1\). The convex face of the lens is directed towards air. In the air, at the distance \(s_1\) from the lens, measured on the principal optic ax, there is a punctual source of monochromatic light.

a) Demonstrate, using Gauss approximation, that between the position of the image, given by the distance \(s_2\) from the lens, and the position of the light source, exists the relation:

\[ \frac{f_1}{s_1}+\frac{f_2}{s_2}=1 \]

where \(f_1\) and \(f_2\) are the focal distances of the lens, in air, respectively in the medium with the refractive index \(n_2\).

Observation: All the refractive indexes are absolute indexes.

b) The lens is cut perpendicular on its plane face in two equal parts. These parts are moved away at a distance \(\delta\ll r\) (Billet lens). On the symmetry axis of the system obtained is led a punctual source of light at the distance \(s_1\) (\(s_1>f_1\)) (fig. 4.1). On the right side of the lens there is a screen E at the distance \(d\). The screen is parallel with the plane face of the lens. On this screen there are \(N\) interference fringes, if on the right side of the lens is air.

Determine \(N\) function of the wave length.

Figure 1

Fig. 4.1

Official English text of the problem.

Answers

  • a) The focal distances and image equation are \[ \boxed{f_1=\frac{R}{n_o-1},\qquad f_2=\frac{n_2R}{n_o-1}}, \qquad \boxed{\frac{f_1}{s_1}+\frac{f_2}{s_2}=1}. \]

  • b) Put \[ f=f_1=\frac{R}{n_o-1},\qquad v=\frac{fs_1}{s_1-f}. \] For the usual screen position, where the outer aperture edges do not limit the interference strip, \[ \boxed{N\simeq N_0 =\frac{\delta^2(s_1+d)} {\lambda\,[d(s_1-f)-fs_1]}}. \] This requires \[ d>v,\qquad r\left(\frac d v-1\right)\ge \delta\left(1+\frac d{s_1}\right). \] Here \(N\) is the usual count obtained by dividing the overlap width by the fringe spacing.

Including finite-aperture clipping, the corresponding count is \[ \boxed{ \overline N= \begin{cases} 0,&d\le v,\\[2mm] \displaystyle \max\!\left\{0,\min\!\left[N_0,\frac{2r\delta}{\lambda f}-N_0\right]\right\}, &d>v. \end{cases}} \] For positive overlap, the integer number of bright maxima, including boundary maxima, is \[ \boxed{N_{\mathrm{bright}} =2\left\lfloor\frac{\overline N}{2}\right\rfloor+1}. \]

Solution

a) Refraction through the two surfaces

Use the thin-lens and Gaussian approximations: neglect the separation of the two surfaces, and retain only first-order terms in ray heights and angles. Take propagation from left to right, with positive \(s_1\) for the real source and positive \(s_2\) for a real image.

Consider a ray incident at height \(y\). Its angle to the optical axis is

\[ \theta_1\simeq\frac{y}{s_1}. \]

Since the centre of curvature is to the right of the convex surface, the surface normal has angle

\[ \nu\simeq-\frac{y}{R}. \]

Let \(\theta_o\) be the ray angle inside the lens. Snell’s law in the small-angle approximation gives

\[ \theta_1-\nu=n_o(\theta_o-\nu), \]

hence

\[ n_o\theta_o =\theta_1-\frac{(n_o-1)y}{R}. \]

At the plane face, whose normal is parallel to the optical axis,

\[ n_2\theta_2=n_o\theta_o. \]

For a ray reaching the axial image,

\[ \theta_2\simeq-\frac{y}{s_2}. \]

Combining these equations and cancelling \(y\) yields

\[ \boxed{\frac1{s_1}+\frac{n_2}{s_2} =\frac{n_o-1}{R}}. \]

The front focal distance is the source distance that makes the outgoing rays parallel. Setting \(s_2\to\infty\),

\[ f_1=\frac{R}{n_o-1}. \]

The rear focal distance is the image distance for parallel incident rays. Setting \(s_1\to\infty\),

\[ f_2=\frac{n_2R}{n_o-1}. \]

Dividing the image equation by \((n_o-1)/R\) therefore gives

\[ \boxed{\frac{f_1}{s_1}+\frac{f_2}{s_2}=1}. \]

The same relation covers virtual images if \(s_2\) is treated as a signed distance.

b) Images formed by the two halves

Both exterior media are now air, so the focal length is

\[ f=\frac{R}{n_o-1}. \]

Each half forms an image at the longitudinal distance \(v\), where

\[ \frac1{s_1}+\frac1v=\frac1f, \qquad v=\frac{fs_1}{s_1-f}>0. \]

Choose the transverse coordinate \(y\) perpendicular to the cut. The full gap shown in the figure is \(\delta\), so the axes of the two halves lie at

\[ c_\pm=\pm\frac{\delta}{2}. \]

Relative to a half-lens axis, the source height is \(-c_\pm\). The transverse magnification is \(-v/s_1\), giving the image positions

\[ y_\pm =c_\pm+\left(-\frac v{s_1}\right)(-c_\pm) =\pm\frac{\delta}{2}\left(1+\frac v{s_1}\right). \]

Consequently, the separation of the two images is

\[ \boxed{b=\delta\left(1+\frac v{s_1}\right) =\frac{\delta v}{f}}. \]

These images act as coherent secondary sources. Reflection symmetry makes their phases equal, so the central fringe is bright.

Fringe spacing

For a screen beyond the image plane, let

\[ L=d-v>0. \]

At screen coordinate \(y\), the paraxial path difference from the two images is

\[ \Delta(y)\simeq\frac{by}{L}. \]

Thus the bright fringes satisfy

\[ y_m=\frac{m\lambda L}{b}, \qquad m\in\mathbb Z, \]

and their spacing is

\[ \boxed{\beta=\frac{\lambda L}{b} =\frac{\lambda f(d-v)}{\delta v}}. \]

They are straight lines parallel to the cut, within the paraxial approximation.

Width of the overlapping beams

We use geometrical beam envelopes and neglect diffraction at the aperture edges.

The ray through the optical centre of each half is undeviated. These two rays reach the screen at

\[ y=\pm H,\qquad H=\frac{\delta}{2}\left(1+\frac d{s_1}\right). \]

Provided the outer aperture edges do not clip the overlap, its width is therefore

\[ W_0=2H=\delta\left(1+\frac d{s_1}\right). \]

Dividing this width by the fringe spacing gives

\[ N_0=\frac{W_0}{\beta} =\frac{\delta^2v}{\lambda f(d-v)} \left(1+\frac d{s_1}\right). \]

Substituting \(v=fs_1/(s_1-f)\),

\[ \boxed{ N_0= \frac{\delta^2(s_1+d)} {\lambda\,[d(s_1-f)-fs_1]} }. \]

For fixed geometry and refractive index, this is proportional to \(1/\lambda\).

Finite-aperture conditions

The statement does not locate the screen relative to the image plane, so the overlap conditions must also be checked.

A ray through height \(y_0\) in the upper half passes through its image at height \(y_+\). Its screen height is

\[ y(d)=\left(1-\frac d v\right)y_0+\frac d v\,y_+. \]

The upper half occupies

\[ \frac{\delta}{2}\le y_0\le\frac{\delta}{2}+r. \]

For \(d>v\), define

\[ \rho=r\left(\frac d v-1\right). \]

On the central section of the screen, the upper beam occupies

\[ H-\rho\le y\le H, \]

and the lower beam occupies

\[ -H\le y\le\rho-H. \]

Their overlap is symmetric about \(y=0\), with width

\[ W= \begin{cases} 0,&\rho\le H,\\ 2(\rho-H),&H<\rho<2H,\\ 2H,&\rho\ge2H. \end{cases} \]

Thus the usual expression \(N_0\) applies when

\[ \rho\ge2H \quad\Longleftrightarrow\quad r\left(\frac d v-1\right) \ge\delta\left(1+\frac d{s_1}\right). \]

In the partially overlapping case,

\[ \frac W\beta =\frac{2\rho}{\beta}-N_0 =\frac{2r\delta}{\lambda f}-N_0. \]

This proves the general expression for \(\overline N=W/\beta\) given in the Answers. For \(d\le v\), the two geometrical beams remain on opposite sides of the symmetry axis and do not overlap.

Finally, when \(W>0\), bright maxima occur at the integers satisfying

\[ |m|\le\frac{W}{2\beta}. \]

Including maxima exactly on the boundary gives

\[ N_{\mathrm{bright}} =2\left\lfloor\frac{W}{2\beta}\right\rfloor+1. \]

The solution was checked against the official solution. Where this solution differs from the official solution, the official solution contains an error or a misprint.