IPhO 1969 · Theory problem 4

Interference in an air layer between a glass plate and a cube

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A glass plate is placed above a glass cube of \(2\ \mathrm{cm}\) edges in such a way that there remains a thin air layer between them, see Figure 5.

Electromagnetic radiation of wavelength between \(400\ \mathrm{nm}\) and \(1150\ \mathrm{nm}\) (for which the plate is penetrable) incident perpendicular to the plate from above is reflected from both air surfaces and interferes. In this range only two wavelengths give maximum reinforcements, one of them is \(\lambda = 400\ \mathrm{nm}\). Find the second wavelength. Determine how it is necessary to warm up the cube so as it would touch the plate. The coefficient of linear thermal expansion is \(\alpha = 8.0 \cdot 10^{-6}\ {}^\circ\mathrm{C}^{-1}\), the refractive index of the air \(n = 1\). The distance of the bottom of the cube from the plate does not change during warming up.

Figure 1

Figure 5:

Official English text of the problem.

Answers

  • Second wavelength: \(\boxed{\lambda_2 \approx 667\ \mathrm{nm}}\).
  • Required warming: increase the cube’s temperature by \(\boxed{\Delta T \approx 3.1\ {}^\circ\mathrm{C}}\). The initial air gap is \(d=500\ \mathrm{nm}\).

Solution

Second wavelength. Let \(d\) be the thickness of the air layer. At its upper boundary, reflection occurs from glass towards air and introduces no phase reversal. At its lower boundary, reflection occurs from air towards glass and introduces a phase reversal of \(\pi\).

At normal incidence, the additional optical path travelled by the second reflected ray is \(2nd\). Thus the phase difference is \[ \Delta\phi=\frac{4\pi nd}{\lambda}+\pi. \] Constructive interference requires an integer multiple of \(2\pi\), giving \[ 2nd=\left(m+\frac12\right)\lambda, \qquad m=0,1,2,\ldots. \] With \(n=1\), the wavelengths of reflection maxima are therefore \[ \lambda_m=\frac{4d}{2m+1}. \]

Let the maximum at \(400\ \mathrm{nm}\) have order \(m\). Then \[ 4d=(2m+1)(400\ \mathrm{nm}). \] Larger orders give wavelengths below \(400\ \mathrm{nm}\), so the second maximum in the specified interval must have order \(m-1\). Consequently \(m\ge1\), and \[ \lambda_{m-1} =400\frac{2m+1}{2m-1}\ \mathrm{nm} \le1150\ \mathrm{nm}. \] Rearrangement gives \[ m\ge\frac{31}{30}, \] so \(m\ge2\).

There must be no third maximum in the interval. The next longer wavelength, corresponding to order \(m-2\), must therefore satisfy \[ \lambda_{m-2} =400\frac{2m+1}{2m-3}\ \mathrm{nm} >1150\ \mathrm{nm}. \] This gives \[ m<\frac{77}{30}. \] The only integer satisfying both bounds is \(m=2\). Hence \[ d=\frac{5(400\ \mathrm{nm})}{4} =\boxed{500\ \mathrm{nm}}, \] and the second wavelength is \[ \boxed{\lambda_2=\frac{4d}{3} =\frac{2000}{3}\ \mathrm{nm} \approx667\ \mathrm{nm}}. \] As a check, the next longer maximum is at \(2000\ \mathrm{nm}\), while the next shorter one is at \(2000/7\approx286\ \mathrm{nm}\); both lie outside the given interval.

Required warming. The cube initially has height \[ h=2\ \mathrm{cm}=2.0\times10^{-2}\ \mathrm{m}. \] Because the distance from its bottom to the plate remains fixed, contact occurs when its height increases by \(d\).

Using linear thermal expansion with the given constant coefficient, \[ \Delta h=\alpha h\,\Delta T=d, \] so \[ \Delta T =\frac{d}{\alpha h} =\frac{5.0\times10^{-7}} {(8.0\times10^{-6})(2.0\times10^{-2})} \ {}^\circ\mathrm{C} =3.125\ {}^\circ\mathrm{C}. \] Thus the required temperature increase is \[ \boxed{\Delta T\approx3.1\ {}^\circ\mathrm{C} \quad\text{(equivalently, }3.1\ \mathrm{K}\text{)}}. \] The relative expansion is only \(d/h=2.5\times10^{-5}\), consistent with the linear approximation.

The solution was checked against the official solution.