IPhO 1974 · Theory problem 2

Plate with a refractive index varying across its width

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Consider a parallel, transparent plate of thickness \(d\) – Fig. 1. Its refraction index varies as

\[ n=\frac{n_0}{1-\dfrac{x}{R}}. \]

Figure 1

Fig. 1

A light beam enters from the air perpendicularly to the plate at the point A (\(x_A=0\)) and emerges from it at the point B at an angle \(\alpha\).

  1. Find the refraction index \(n_B\) at the point B.
  2. Find \(x_B\) (i.e. value of \(x\) at the point B)
  3. Find the thickness \(d\) of the plate.

Data:

\[ n_0=1.2;\qquad R=13\,\mathrm{cm};\qquad \alpha=30^\circ. \]

Official English text of the problem.

Answers

  1. \(\displaystyle n_B=\sqrt{n_0^2+\sin^2\alpha}=\boxed{1.3}\) (dimensionless).
  2. \(\displaystyle x_B=R\left(1-\frac{n_0}{\sqrt{n_0^2+\sin^2\alpha}}\right)=\boxed{1.0\,\mathrm{cm}}\).
  3. \(\displaystyle d=\frac{R\sin\alpha}{\sqrt{n_0^2+\sin^2\alpha}}=\boxed{5.0\,\mathrm{cm}}\).

Solution

1. Refraction index at B

We use geometrical optics, treating the refractive index as varying smoothly on the scale of the wavelength, and take the refractive index of air to be \(1\).

Let \(\beta\) denote the angle the ray inside the plate makes with the positive \(y\)-axis, which is normal to the plate surfaces. Normal incidence at A gives \(\beta_A=0\).

To determine how the ray bends, imagine dividing the plate into infinitesimally thin vertical layers of constant refractive index. The normal to each layer interface points along \(x\), so the ray’s angle to that normal is \(90^\circ-\beta\). Snell’s law across successive layers therefore gives

\[ n\sin(90^\circ-\beta)=n\cos\beta=\text{constant}. \]

Evaluating this constant at A, where \(n=n_0\), yields

\[ \boxed{n(x)\cos\beta=n_0}. \]

At B, let the internal incidence angle be \(\beta_B\). Thus

\[ n_B\cos\beta_B=n_0. \]

Snell’s law at the upper surface, where the ray emerges into air, gives

\[ n_B\sin\beta_B=\sin\alpha. \]

Squaring and adding these equations eliminates \(\beta_B\):

\[ n_B^2(\cos^2\beta_B+\sin^2\beta_B) =n_0^2+\sin^2\alpha. \]

Hence

\[ \boxed{n_B=\sqrt{n_0^2+\sin^2\alpha}}. \]

Numerically,

\[ n_B=\sqrt{1.2^2+\sin^2 30^\circ} =\sqrt{1.44+0.25} =\boxed{1.3}. \]

2. Position \(x_B\)

Using the prescribed index profile at B,

\[ n_B=\frac{n_0}{1-x_B/R}, \]

we obtain

\[ \boxed{x_B=R\left(1-\frac{n_0}{n_B}\right)}. \]

Substitution gives

\[ x_B=13\,\mathrm{cm}\left(1-\frac{1.2}{1.3}\right) =13\,\mathrm{cm}\left(\frac{1}{13}\right) =\boxed{1.0\,\mathrm{cm}}. \]

3. Thickness \(d\)

Combining the conserved quantity with the index profile gives

\[ \cos\beta=\frac{n_0}{n(x)}=1-\frac{x}{R}. \]

The refractive index increases with \(x\), so the ray bends towards positive \(x\). On this branch, its tangent satisfies

\[ \frac{dy}{dx}=\cot\beta =\frac{\cos\beta}{\sqrt{1-\cos^2\beta}} =\frac{R-x}{\sqrt{2Rx-x^2}}. \]

Integrating from A, where \(x=y=0\), to an arbitrary point on the ray,

\[ y(x)=\int_0^x\frac{R-u}{\sqrt{2Ru-u^2}}\,du =\sqrt{2Rx-x^2}. \]

Consequently, the ray follows the circular arc

\[ \boxed{(x-R)^2+y^2=R^2}, \]

with centre \((R,0)\) and radius \(R\). The relevant arc starts at A with a vertical tangent and bends to the right.

At B, \(y=d\), so

\[ d=\sqrt{2Rx_B-x_B^2}. \]

Equivalently, substituting the expression for \(x_B\),

\[ d=R\sqrt{1-\left(\frac{n_0}{n_B}\right)^2} =\boxed{\frac{R\sin\alpha}{\sqrt{n_0^2+\sin^2\alpha}}}. \]

Numerically,

\[ d=\sqrt{2(13\,\mathrm{cm})(1.0\,\mathrm{cm}) -(1.0\,\mathrm{cm})^2} =\sqrt{25\,\mathrm{cm}^2} =\boxed{5.0\,\mathrm{cm}}. \]

The solution was checked against the official solution.