IPhO 1971 · Theory problem 4

Fish in a spherical aquarium in front of a plane mirror

Download PDFIPhO 1971

A spherical aquarium, filled with water, is placed in front of a flat vertical mirror. The radius of the aquarium is \(R\), and the distance between its center and the mirror is \(3R\). A small fish, which is initially at the point nearest to the mirror, starts to move with velocity \(v\) along the wall. An observer looks at the fish from a very large distance along a horizontal line passing trough the center of the aquarium.

What is the relative velocity \(v_{\mathrm{rel}}\) at which the two images of the fish seen by the observer will move apart? Express your answer in terms of \(v\). Assume that:

  • The wall of the aquarium is made of a very thin glass.
  • The index of refraction of water is \(n=4/3\).

Official English text of the problem.

Answers

  • 4. The initial relative velocity is \[ \boxed{v_{\mathrm{rel}}=\frac{8}{3}v}. \] The direct image moves at \(2v\), while the image seen through reflection moves in the opposite direction at \(2v/3\).

Solution

4. Take the viewing axis as the \(x\)-axis, directed from the observer toward the mirror. Put the aquarium centre at \(x=0\), the mirror at \(x=3R\), and the fish initially at \(x=R\).

Choose the transverse \(y\)-direction along the fish’s initial motion. Because the fish moves tangentially to the spherical wall, its displacement to first order in time is \[ y=vt, \] while its axial displacement is of second order. Thus we may keep its axial coordinate equal to \(R\) when calculating the initial image velocities.

The thin glass can be neglected: refractions at its two nearly coincident surfaces combine into refraction directly between water and air. We use paraxial Snell’s law, which is sufficient for the initial motion about the axis.

By reversibility of light, trace rays backward from the distant observer toward the fish. To the required accuracy, these rays initially travel parallel to the axis. Their height \(h\) at the observer-facing surface, \(x=-R\), gives the apparent transverse image position.

At this surface, the normal directed into the water makes the small angle \(-h/R\) with the axis. If the refracted ray makes angle \(\alpha\), Snell’s law gives \[ 0+\frac{h}{R} =n\left(\alpha+\frac{h}{R}\right). \] Consequently, \[ \alpha=-\frac{n-1}{n}\frac{h}{R} =-\frac{h}{4R}. \] After crossing the aquarium, its height at the mirror-facing surface, \(x=R\), is \[ k=h+2R\alpha=\frac{h}{2}. \]

Direct image. For the direct path, the backward-traced ray reaches the fish at this surface, so \(k=y\). Therefore \[ h_{\mathrm d}=2y, \qquad \boxed{\dot h_{\mathrm d}=2v}. \]

Image seen through reflection. For this path, the backward-traced ray continues through the mirror-facing surface toward the mirror. The outward surface normal makes angle \(k/R\) with the axis. If the ray’s angle in air is \(\gamma\), Snell’s law gives \[ n\left(\alpha-\frac{k}{R}\right) =\gamma-\frac{k}{R}. \] Using \(n=4/3\), \(\alpha=-h/(4R)\), and \(k=h/2\), \[ \gamma=n\alpha-(n-1)\frac{k}{R} =-\frac{h}{2R}. \]

We can unfold the plane-mirror reflection by replacing the fish with its mirror image. Its axial coordinate is \[ x_{\mathrm{mirror\ image}}=2(3R)-R=5R, \] and its transverse coordinate remains \(y\). Since the fish is at the wall, refraction where the ray finally reaches it introduces no additional propagation distance.

The unfolded ray travels an axial distance \(4R\) from the aquarium’s mirror-facing surface to this mirrored position. Hence \[ y=k+4R\gamma =\frac{h}{2}+4R\left(-\frac{h}{2R}\right) =-\frac{3h}{2}. \] Thus the reflected image has apparent transverse position and velocity \[ h_{\mathrm r}=-\frac{2}{3}y, \qquad \boxed{\dot h_{\mathrm r}=-\frac{2}{3}v}. \]

The two images move in opposite transverse directions. Their initial relative velocity is therefore \[ \boxed{ v_{\mathrm{rel}} =\left|\dot h_{\mathrm d}-\dot h_{\mathrm r}\right| =2v+\frac{2}{3}v =\frac{8}{3}v }. \]

The solution was checked against the official solution.