IPhO 1983 · Theory problem 3

Prisms

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Two dispersive prisms having apex angles \(\hat A_1=60^\circ\) and \(\hat A_2=30^\circ\) are glued as in the figure below \((\hat C=90^\circ)\). The dependences of refraction indexes of the prisms on the wavelength are given by the relations

\[ n_1(\lambda)=a_1+\frac{b_1}{\lambda^2}; \]

\[ n_2(\lambda)=a_2+\frac{b_2}{\lambda^2} \]

were

\[ a_1=1{,}1,\quad b_1=1\cdot10^5\,nm^2,\quad a_2=1{,}3,\quad b_2=5\cdot10^4\,nm^2. \]

Figure 1

a. Determine the wavelength \(\lambda_0\) of the incident radiation that pass through the prisms without refraction on \(AC\) face at any incident angle; determine the corresponding refraction indexes of the prisms.

b. Draw the ray path in the system of prisms for three different radiations \(\lambda_{red},\lambda_0,\lambda_{violet}\) incident on the system at the same angle.

c. Determine the minimum deviation angle in the system for a ray having the wavelength \(\lambda_0\).

d. Calculate the wavelength of the ray that penetrates and exits the system along directions parallel to \(DC\).

Official English text of the problem.

Answers

  • a. \(\boxed{\lambda_0=500\,\mathrm{nm}}\), with \(\boxed{n_1(\lambda_0)=n_2(\lambda_0)=1.5}\).
  • b. At \(AC\), red bends toward the normal, \(\lambda_0\) continues straight, and violet bends away from the normal. For a common horizontal incident beam, the emerging rays are ordered red, \(\lambda_0\), violet from top to bottom, with increasing downward deviation.
  • c. \(\boxed{\delta_{\min}=2\arcsin(1.5\sin15^\circ)-30^\circ\approx15.69^\circ}\).
  • d. \(\boxed{\lambda\approx1.18\times10^3\,\mathrm{nm}=1.18\,\mu\mathrm m}\), in the infrared.

Solution

We use geometrical optics and Snell’s law, taking the refractive index of the surrounding air as \(1\) and neglecting any optical effect of the glue. No small-angle approximation is used.

The geometry gives \[ \angle ACD=90^\circ-A_2=60^\circ, \qquad \angle ADC=180^\circ-A_1-\angle ACD=60^\circ. \] Consequently, measured relative to the rightward direction parallel to \(DC\):

  • the inward normal to \(AD\) points \(30^\circ\) downward;
  • the normal to \(AC\), directed from prism 1 into prism 2, points \(30^\circ\) upward;
  • the outward normal to \(BC\) is horizontal.

a. Wavelength with no refraction at \(AC\)

At the common face, Snell’s law is \[ n_1(\lambda)\sin i=n_2(\lambda)\sin t. \] For the direction to remain unchanged, \(t=i\). Requiring this for every incidence angle, rather than just normal incidence, gives \[ n_1(\lambda_0)=n_2(\lambda_0). \] Thus \[ a_1+\frac{b_1}{\lambda_0^2} =a_2+\frac{b_2}{\lambda_0^2}, \] and therefore \[ \lambda_0^2 =\frac{b_1-b_2}{a_2-a_1} =\frac{5\times10^4\,\mathrm{nm}^2}{0.2} =2.5\times10^5\,\mathrm{nm}^2. \] Taking the positive wavelength, \[ \boxed{\lambda_0=500\,\mathrm{nm}}. \]

The corresponding indices are \[ n_1(\lambda_0)=1.1+\frac{10^5}{500^2}=1.5, \qquad n_2(\lambda_0)=1.3+\frac{5\times10^4}{500^2}=1.5. \]

b. Ray paths

Both indices decrease with increasing wavelength. Hence, at the entrance face \(AD\), violet bends toward the normal more strongly than \(\lambda_0\), while red bends less strongly.

At the common face, the relevant index difference is \[ n_1(\lambda)-n_2(\lambda) =-0.2+\frac{5\times10^4\,\mathrm{nm}^2}{\lambda^2}. \] Since \[ \lambda_{\mathrm{violet}}<500\,\mathrm{nm}<\lambda_{\mathrm{red}}, \] the refractions at \(AC\) are:

Radiation Index comparison Behaviour at \(AC\)
Red \(n_1<n_2\) Bends toward the normal
\(\lambda_0\) \(n_1=n_2\) Continues straight
Violet \(n_1>n_2\) Bends away from the normal

A precise verbal sketch can use a common horizontal incident beam, whose incidence angle at \(AD\) is \(30^\circ\). Choose its entry point sufficiently high that all three rays reach \(BC\).

  1. At \(AD\), all three rays bend downward toward the inward normal. Inside prism 1, red is the upper ray, \(\lambda_0\) the middle ray, and violet the lower ray.
  2. At \(AC\), draw its normal pointing \(30^\circ\) above horizontal. Red turns upward toward this normal, \(\lambda_0\) remains straight, and violet turns farther downward away from the normal.
  3. At \(BC\), the normal is horizontal. The visible rays bend farther downward as they pass into air, away from this normal.

The emerging order is therefore red above \(\lambda_0\), with violet below, and violet has the greatest total deviation.

c. Minimum deviation for \(\lambda_0\)

For this wavelength, both prisms have index \(n=1.5\), so the internal face \(AC\) has no optical effect. The entrance and exit faces \(AD\) and \(BC\) act as the faces of a single prism with effective apex angle \[ \alpha=A_1-A_2=30^\circ. \]

Let \(i,e\) be the external incidence and emergence angles, and \(r,r'\) the corresponding internal angles. Prism geometry and Snell’s law give \[ r+r'=\alpha,\qquad \delta=i+e-\alpha, \] \[ \sin i=n\sin r,\qquad \sin e=n\sin r'. \] Therefore \[ \delta(r)= \arcsin(n\sin r) +\arcsin\!\bigl(n\sin(\alpha-r)\bigr)-\alpha. \]

The minimum occurs for the symmetric path, \[ r=r'=\frac{\alpha}{2},\qquad i=e. \] For example, differentiating the preceding expression gives \[ \frac{d\delta}{dr} = \frac{n\cos r}{\sqrt{1-n^2\sin^2r}} - \frac{n\cos(\alpha-r)} {\sqrt{1-n^2\sin^2(\alpha-r)}}. \] This vanishes at \(r=\alpha/2\) and changes from negative to positive there.

Thus \[ i_{\min} =\arcsin\!\left(n\sin\frac{\alpha}{2}\right) =\arcsin(1.5\sin15^\circ) \approx22.8443^\circ, \] and \[ \boxed{\delta_{\min} =2i_{\min}-\alpha \approx15.6886^\circ \approx15.69^\circ}. \]

d. Parallel incident and emerging rays

The incident ray is horizontal, so its incidence angle at \(AD\) is \(30^\circ\). If its refracted angle there is \(r\), Snell’s law gives \[ n_1\sin r=\sin30^\circ=\frac12. \tag{1} \]

The emerging ray must also be horizontal. Since \(BC\) is vertical, emergence is normal to \(BC\). Snell’s law then requires the ray inside prism 2 to be horizontal as well.

The ray in prism 1 travels at angle \(r-30^\circ\) to the horizontal. Its incidence angle at \(AC\) is consequently \[ i_{AC}=30^\circ-(r-30^\circ)=60^\circ-r. \] The horizontal ray in prism 2 makes \(30^\circ\) with the normal to \(AC\). Hence \[ n_1\sin(60^\circ-r)=n_2\sin30^\circ=\frac{n_2}{2}. \tag{2} \]

Expanding equation (2) and using equation (1), \[ n_2=\sqrt3\,n_1\cos r-n_1\sin r =\sqrt3\sqrt{n_1^2-\frac14}-\frac12. \] Squaring and simplifying gives \[ \boxed{3n_1^2=n_2^2+n_2+1}. \tag{3} \]

Introduce the positive, dimensionless quantity \[ u=\frac{5\times10^4\,\mathrm{nm}^2}{\lambda^2}. \] Then \[ n_1=1.1+2u,\qquad n_2=1.3+u. \] Substitution into equation (3) yields \[ 3(1.1+2u)^2=(1.3+u)^2+(1.3+u)+1, \] or \[ 11u^2+9.6u-0.36=0. \] Only the positive root is physical: \[ u=\frac{-9.6+\sqrt{108}}{22} \approx0.0360139. \] Therefore \[ \lambda =\sqrt{\frac{5\times10^4\,\mathrm{nm}^2}{u}} \approx1178.28\,\mathrm{nm}, \] so \[ \boxed{\lambda\approx1.18\times10^3\,\mathrm{nm} =1.18\,\mu\mathrm m}. \]

As a check, this wavelength gives \[ n_1\approx1.17203,\qquad n_2\approx1.33601, \] and \(r\approx25.2527^\circ\). The incidence angle at \(AC\) is then \(34.7473^\circ\), which refracts to exactly \(30^\circ\); the ray subsequently reaches \(BC\) normally and exits horizontally.

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.