Ray paths in a Wollaston prism
As a reminder, we can recall that a Wollaston prism is made of two right-angle prisms cut in the same uniaxial birefringent crystal, generally a quartz crystal. The two prisms have parallel cuts and the two optical axes are perpendicular. To make the exercise simpler, we will only consider the rays for which the plane \((x , z )\) perpendicular to the edges of the Wollaston prism is the plane of incidence.
Question
Using Huygens' construction, explain the ordinary and extraordinary ray paths in Wollaston prism for a null angle of incidence.
Solution
The direction \(z\) corresponds to the direction of propagation of the light and \((x , z )\) is the plane of incidence. On the following animation, the first prism reached by the light is green with a purple checkerboard, and the second is blue. In I three wave surfaces are represented: the first one is purple and of spherical shape, and corresponds to the ordinary wave surface of the two prisms; the second one is of a gray-green color and is relative to the extraordinary wave surface of the first prism; finally the third one is red and shows the extraordinary wave surface of the second prism. Quartz being a positive crystal, the ellipsoidal wave surfaces are inside the spherical wave surface.
The wave surface corresponding to the ordinary ray in the first prism is tangent to the gray-green ellipsoid relative to the extraordinary wave surface, the tangent direction being parallel to the direction of the optical axis, which direction is the same as the axis \(x\). This ordinary wave surface is also tangent to the red ellipsoid relative to the extraordinary wave surface in the second prism, the tangent direction being now parallel to the axis \(y\).
The animation shows these wave surfaces under different angles of view. The first angle shot corresponds to Figure 2, then the camera moves to face the plane \((y , z )\) and finally it moves upward to face the plane of incidence \((x , z )\). Figure 3 corresponds to the animation's last angle shot.
Figure 3 shows these wave surfaces, the prisms being seen from above and following direction \(y\). See Figure 4 for a configuration of Figure 3.
From now, we will be thinking from Figure 4. Since the angle of incidence is nil at the level of the first diopter, the angle of refraction has a zero value as well, and in the first prism the ordinary and extraordinary rays are not separated, but they propagate at different speeds. The thickness of the glue between the two prisms acts as a parallel faces plate of very little thickness. We decide to ignore the slight translation of the ray due to this plate.
The axes of the two prisms being crossed, the ordinary ray in the first prism becomes extraordinary in the second and vice versa. The second animation shows the propagation of the electric field vector. The incident light is polarized with a linear polarization and a direction at a \(45^{\circ}\) angle to the directions of the neutral lines in the two prisms' quartz crystals. Both components of the electric field vector are directed following these neutral lines. These two components are depicted in black when they are in the air.
When going through a diopter, concentric circles have been depicted, in order to get a visual representation of the phenomenon, but they do not correspond to any physical phenomenon.
In the first prism, the component depicted in red is relative to the extraordinary ray, and the blue one to the ordinary ray. The latter propagates faster, because the crystal is positive.
When passing through the second prism, the ordinary and extraordinary rays exchange their functions for the crossed crystals' axes.
In the air the two components go back to propagating at the same speed.
For the first prism, the main plane is contained in the plane of incidence which is the same as the plane of Figure 4. The wave surfaces relative to the ordinary and extraordinary rays are noted \(S_{o,1}\) and \(S_{e,1}\) . For the second prism the spherical wave surface \(S_{o,2}\) relative to the ordinary beam is not separated from \(S_{o,1}\) since both prisms are made from the same material. Conversely for the extraordinary wave surface \(S_{e,2}\),the main axis of the ellipse is perpendicular to the figure's plane, since the optical axis is perpendicular to the plane of incidence.
The prolongation of the ordinary incident ray touches the ordinary wave surface at point O. The wave plane \(P_{o,1}\), tangent in O, touches the separating diopter of the prisms in K. During the refraction, no phase difference is introduced, the refracted wave plane also passes through point K. This wave plane \(P_{o,2}\) is relative to the extraordinary ray, since as we have already mentioned it, the axes of both prisms being crossed, the ordinary ray in the first prisms becomes extraordinary; this plane is then tangent in a point A at the extraordinary wave surface \(S_{o,2}\) of the second prism. The extraordinary ray \(R_{e}\)passes by the point A. At the level of the diopter between the second prism and the air, Descartes Laws simply apply :
\(n_{e} \sin i_{e} = n \sin I_{e}\)
\(n\) being the index of the outside environment, which is generally air.
In the same way, the extraordinary incident ray's prolongation touches the extraordinary wave's surface in a point E. The wave plane \(P_{e,1}\) ,tangent in E, touches the prisms' separating diopter in K'. During the refraction, no phase difference is introduced, the refracted wave plane passes through the point K'. This wave plane \(P_{o,2}\) is relative to the ordinary ray; the axes of both prisms being crossed (the extraordinary ray in the first prism becomes ordinary), this plane is tangent in a point B to the ordinary wave surface \(S_{o,2}\) of the second prism. The ordinary ray \(R_{o}\) passes by point B. Out of the second prism, Descartes Laws apply again:
\(n_{o} \sin i_{o} = n \sin I_{o}\)
\(n\) being the air index.
Remarque :
This type of construction gives access to the rays direction, that is to say to the energy flows (direction of Poynting vectors), but not to the wave vectors' direction. It corresponds to the geometrical optical classic constructions which indicate the energy's path. However, in this particular case - the incidence being normal and the Wollaston prism being crossed in this direction - there is no difference between the Poynting vectors' directions and the wave vectors' directions in both prisms and in the air (one of the reasons for this is that the cut of the ellipsoid with the plane of incidence happens to be a circle in the second prism).
Question
Show that the angle formed between the ordinary and the extraordinary rays is roughly worth : \(2(n_{e} - n_{o}) \tan \alpha\)
Solution
At the level of the separating diopter between the two prisms, the angle of incidence of the ordinary and extraordinary rays is equal to the angle of the prism \(\alpha\). Applying Descartes Law at a point I:
\(n_{o} \sin ( \alpha ) = n_{e} \sin ( \alpha - i_{e} ) = n_{e} [ \sin ( \alpha ) \cdot \cos ( i_{e} ) - \cos ( \alpha ) \cdot \sin ( i_{e} )]\)
\(n_{e} \sin ( \alpha ) = n_{o} \sin ( \alpha + i_{o} ) = n_{o} [ \sin ( \alpha ) \cdot \cos ( i_{o} ) + \cos ( \alpha ) \cdot \sin ( i_{o} )]\)
Angles \(i_{e}\) and \(i_{o}\) being small:
\(n_{o} \sin ( \alpha ) \approx n_{e} \cdot \sin ( \alpha ) - \cos ( \alpha ) \cdot n_{e} i_{e}\)
\(n_{e} \sin ( \alpha ) \approx n_{o} \cdot \sin ( \alpha ) + \cos ( \alpha ) \cdot n_{o} i_{o}\)
Hence :
\(\sin ( \alpha ) [n_{e} - n_{o}] = n_{e}i_{e} \cdot \cos ( \alpha )\)
\(\sin ( \alpha ) [n_{e} - n_{o}] = n_{o}i_{o} \cdot \cos ( \alpha )\)
And :
\(n_{e} i_{e} = (n_{e} - n_{o}) \cdot \tan ( \alpha )\)
\(n_{o} i_{o} = (n_{e} - n_{o}) \cdot \tan ( \alpha )\)
Out of the second prism in the air, at the level of the third diopter and considering that the index \(n\) of the air is equal to 1:
\(n_{e} \sin ( i_{e}) = \sin ( I_{e})\)
\(n_{o} \sin ( i_{o}) = \sin ( I_{o})\)
Considering the approximation of the small angles:
\(\sin (I_{e}) \approx n_{e} i_{e} = (n_{e} - n_{o}) \cdot \tan ( \alpha ) \approx I_{e}\)
\(\sin (I_{o}) \approx n_{o} i_{o} = (n_{e} - n_{o}) \cdot \tan ( \alpha ) \approx I_{o}\)
The value of the angle between the two rays is equal to:
\(I_{e} + I_{o} =2(n_{e}-n_{o}) \cdot \tan ( \alpha ) = \in\)
Question
This Wollaston prism is used with a infinity-corrected objective of a \(4 mm\) focal (\(\times 50\) magnification associated with a lens of a \(200mm\) tube) and a \(0.8\) numerical aperture. What must the value of angle \(\alpha\) be if we want the ordinary and extraordinary rays to be separated from one another by an angle equal to \(2/3\) of the resolution of the objective in its image space, that is to say, according to Rayleigh criterion for an objective limited by the diffraction, to \(2/3\) of the angular radius of the Airy disk? (the ordinary and extraordinary indexes of the quartz in the visible are \(n_{o} \approx 1.544\) and \(n_{e} \approx 1.553\)).
To solve this problem, the lesson on birefringent medium must be referred to ([ Surrel[1]] for a simplified approach and [ Huard[2]] to take it further .
Solution
The value of the radius of the Airy disk due to the objective in its object space is equal to:
\(r = 1.22 \frac{\lambda}{2 \cdot NA} = 1.22 \cdot \frac{0.55}{2 \cdot 0.80} = 0.42 \mu m\)
which corresponds to an angular radius in its image space of:
\(\frac{r}{f'} \approx 10^{-4} rad\)
which therefore leads to a necessary angular deviation between the two ordinary and extraordinary rays of:
\(\epsilon = \frac{2}{3} \frac{r}{f'} \approx 0.7 \cdot 10^{-4}rad\)
hence an angle for the prism of:
\(\alpha = \arctan \bigl\{ \frac{\epsilon}{2(n_{e}-n_{o}} \bigl\} \approx 55 \cdot 10^{-4}rad \approx 0.2^{\circ}\)