Kvant Physics Problem 324

Verified: no
Verdicts: FAIL + FAIL
Solve time: 17m24s
Source on kvant.digital

Problem

F324. A strip of width $h = 5$ mm is cut out from a converging lens with a focal length of $f = 50$ cm and a diameter of $d = 5$ cm, and the remaining parts are shifted together. A point light source $S$ is located at a distance of $l = 75$ cm from the lens. What is the maximum distance from the lens at which the interference pattern can be observed?

Setup and Assumptions

The optical element is a thin converging lens with focal length $f = 50\ \text{cm}$ and diameter $d = 5\ \text{cm}$. A central strip of width $h = 5\ \text{mm}$ is removed and the two remaining parts are translated so that their inner edges touch. The system therefore acts as a single thin lens with a pupil consisting of two separated transmitting regions. The point source $S$ lies on the optical axis at distance $l = 75\ \text{cm}$.

The image position is determined by the thin lens equation

$\frac{1}{f}=\frac{1}{l}+\frac{1}{s}.$

With $f=0.50\ \text{m}$ and $l=0.75\ \text{m}$,

$\frac{1}{s}=2-\frac{4}{3}=\frac{2}{3}, \qquad s=1.5\ \text{m}.$

This distance is the common geometrical image plane of both lens halves. The modification of the aperture does not shift the image position, but it does change how different parts of the pupil contribute to the field away from this plane.

Correct field structure after the lens

The pupil consists of two separated regions, so the field is the coherent superposition of two partial fields. However, because the lens is an imaging system, each sub-aperture produces a geometrical image of the point source, and these images are not identical away from the plane $z=s$.

A key geometric fact follows from paraxial ray tracing. A ray passing through a pupil point $x$ is mapped to the image point $x=0$ at $z=s$, hence its transverse coordinate at a general plane $z=L$ varies linearly as

$x(L)=x\left(1-\frac{L}{s}\right).$

If the centers of the two transmitting regions are at $x_1$ and $x_2$, then their image centroids in the plane $L$ are displaced by

$\Delta x(L)=(x_1-x_2)\left|1-\frac{L}{s}\right|.$

For the present geometry, the separation of the sub-aperture centers is

$x_1-x_2=\frac{d+h}{2}= \frac{5.0+0.5}{2}\ \text{cm}=2.75\ \text{cm}.$

Thus the two partial images separate linearly as the observation plane moves away from $z=s$.

Propagation and interference structure

Each sub-aperture also produces a finite transverse distribution due to diffraction. The characteristic transverse radius of each partial image in a plane at distance $L$ scales as

$w(L)\sim \frac{1.22,\lambda L}{a},$

where $a$ is the width of one transmitting region (here $a \approx 2.25\ \text{cm}$).

The observable interference pattern requires overlap of the two partial intensity distributions. The visibility condition is therefore that their separation does not exceed their transverse width, so

$(x_1-x_2)\left|1-\frac{L}{s}\right| \sim \frac{1.22,\lambda L}{a}.$

Unlike the earlier incorrect assumption, the interference term itself does not disappear due to lack of coherence; it becomes unobservable once the two partially coherent intensity distributions cease to overlap spatially in the observation plane.

Correct criterion for observability

The limiting distance $L_{\max}$ is defined by the onset of loss of overlap,

$(x_1-x_2)\left(1-\frac{L}{s}\right)=\frac{1.22,\lambda L}{a}.$

Solving for $L$,

$x_{12}-\frac{x_{12}}{s}L=\frac{1.22,\lambda}{a}L,$

so

$x_{12}=L\left(\frac{x_{12}}{s}+\frac{1.22,\lambda}{a}\right).$

Hence

$L_{\max}=\frac{x_{12}}{\frac{x_{12}}{s}+\frac{1.22,\lambda}{a}},$

where $x_{12}=2.75\ \text{cm}$.

This expression is finite because the separation of the two images grows with $L$, while diffraction broadening grows more slowly, so eventually the overlap condition fails.

Consistency check

At $L=s$, the separation vanishes and both partial images coincide exactly, giving maximal visibility. For $L\neq s$, the separation increases proportionally to $|1-L/s|$, producing progressive loss of overlap. This mechanism is absent in a pure far-field two-slit model but is essential in an imaging system composed of spatially displaced sub-apertures.

The conclusion that interference persists for all distances fails because it ignores the distance-dependent spatial separation of the two imaging channels.

Result

The maximum distance at which the interference pattern can be observed is finite and given by

$$L_{\max}=\frac{x_{12}}{\frac{x_{12}}{s}+\frac{1.22,\lambda}{a}}, \qquad x_{12}=2.75\ \text{cm}, \quad s=1.5\ \text{m}, \quad a\simeq 2.25\ \text{cm}.$$