IPhO 1970 · Theory problem 4

Spherical telescope mirror and a disk receiver in its focus

Download PDFIPhO 1970

A spherical mirror is installed into a telescope. Its lateral diameter is \(D=0{,}5\ \mathrm{m}\) and the radius of the curvature \(R=2\ \mathrm{m}\). In the main focus of the mirror there is an emission receiver in the form of a round disk. The disk is placed perpendicular to the optical axis of the mirror (Fig. 4). What should the radius \(r\) of the receiver be so that it could receive the entire flux of the emission reflected by the mirror? How would the received flux of the emission decrease if the detector’s dimensions decreased by 8 times?

Directions: 1) When calculating small values \(\alpha\) (\(\alpha\ll1\)) one may perform a substitution \(\sqrt{1-\alpha}\approx1-\dfrac{\alpha}{2}\); 2) diffraction should not be taken into account.

Figure 1

Fig. 4

Official English text of the problem.

Answers

  • Required receiver radius: \[ \boxed{r_{\min}\simeq\frac{D^3}{16R^2}\approx2.0\ \mathrm{mm}.} \] Without the small-aperture expansion, the geometrical result is \(2.024\ldots\ \mathrm{mm}\).

  • Flux after an eightfold reduction in linear dimensions: \[ \boxed{\frac{\Phi_{\mathrm{small}}}{\Phi_{\mathrm{full}}}\simeq\frac14.} \] The received flux decreases by approximately a factor of 4, corresponding to a 75% loss. Unexpanded ray geometry gives a retained fraction of \(0.25449\ldots\).

Solution

Required receiver radius

Assume a distant source on the optical axis, providing parallel rays and uniform illumination across the mirror’s aperture. We use geometrical optics and neglect diffraction.

Let the mirror’s vertex be at \(z=0\), with its centre of curvature at \(z=R\). Its paraxial focus is therefore at \[ f=\frac R2=1\ \mathrm{m}. \] Write the aperture radius as \[ a=\frac D2=0.25\ \mathrm{m}. \]

Consider an incident ray at height \(h\) above the axis. At its point of reflection \(P\), let the normal make an angle \(\theta\) with the axis. The spherical surface gives \[ h=R\sin\theta,\qquad z_P=R(1-\cos\theta). \]

By the law of reflection, the reflected ray travels towards the axis at angle \(2\theta\). Its signed height when it reaches the focal plane \(z=f\) is \[ y_F=h-(f-z_P)\tan2\theta. \] Substituting the expressions above, \[ \begin{aligned} y_F &=R\sin\theta -R\left(\cos\theta-\frac12\right)\tan2\theta\\ &=-\frac{R\sin\theta(1-\cos\theta)}{\cos2\theta}. \end{aligned} \] The negative sign means that the ray has already crossed the axis before reaching the paraxial focal plane: this is spherical aberration.

Thus its distance from the axis in that plane is \[ \boxed{\rho(h)= \frac{h\left(1-\sqrt{1-h^2/R^2}\right)} {1-2h^2/R^2}.} \] For the present aperture, the numerator increases with \(h\), while the positive denominator decreases. Hence \(\rho(h)\) increases, and the outermost rays determine the required disk radius: \[ r_{\min}=\rho(a) =\frac{a\left(1-\sqrt{1-a^2/R^2}\right)} {1-2a^2/R^2}. \] With the given values, this is \[ r_{\min}=2.024\ldots\times10^{-3}\ \mathrm{m}. \]

To obtain the requested small-quantity approximation, note that \[ \frac{h^2}{R^2}\leq\frac{a^2}{R^2}=\frac1{64}\ll1. \] Using the supplied expansion, \[ 1-\sqrt{1-\frac{h^2}{R^2}}\simeq\frac{h^2}{2R^2}. \] Keeping the leading nonzero term in \(\rho(h)\) therefore gives \[ \rho(h)\simeq\frac{h^3}{2R^2}. \] Consequently, \[ \boxed{ r_{\min}\simeq\frac{a^3}{2R^2} =\frac{D^3}{16R^2} =\frac{(0.5)^3}{16(2)^2}\ \mathrm{m} =1.95\times10^{-3}\ \mathrm{m} \approx2.0\ \mathrm{mm}. } \]

Flux after reducing the detector’s dimensions

An eightfold reduction in linear dimensions makes the detector radius \[ r_{\mathrm{small}}=\frac{r_{\min}}8. \]

Since \(\rho(h)\) increases with \(h\), this smaller disk receives the rays reflected from a central portion of the mirror’s aperture, of radius \(b\), defined by \[ \rho(b)=\frac{\rho(a)}8. \] Using the cubic dependence just derived, \[ \frac{b^3}{2R^2}=\frac18\frac{a^3}{2R^2}, \] so \[ b=\frac a2. \]

For uniform incident illumination and uniform reflectivity, the reflected flux is proportional to the projected aperture area contributing to the detector. Therefore, \[ \boxed{ \frac{\Phi_{\mathrm{small}}}{\Phi_{\mathrm{full}}} =\frac{\pi b^2}{\pi a^2} \simeq\left(\frac12\right)^2 =\frac14. } \] Thus the received flux decreases approximately fourfold.

For comparison, using the unexpanded expression for \(\rho(h)\), the equation \(\rho(b)=\rho(a)/8\) gives \(b=0.126117\ldots\ \mathrm{m}\), and hence \[ \frac{\Phi_{\mathrm{small}}}{\Phi_{\mathrm{full}}} =0.25449\ldots, \] consistent with the leading-order result.

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.