14.8 In the Fig. 14.28 \(P_1\) represents a polarizer whose transmission axis forms an angle of \(30^{\circ }\) with respect to the vertical axis OX . L is a plate of a uniaxial crystal with thickness d unknown. The optic axis is parallel to its faces. \(P_2\) is an analyzer with a transmission axis that makes an angle \(\varphi \) with respect to the vertical axis. The system is illuminated perpendicularly with a radiation composed by two wavelengths of \(\lambda _1=400\) nm and \(\lambda _1=600\) nm, respectively. Calculate the minimum thickness d of the plate up to \(1\,\) mm so that the analyzer \(P_2\) is able to completely avoid that one of the wavelengths crosses \(P_2\) . Explain the value of \(\varphi \) for all the solutions found. Assume that \(n_1-n_2=0.01\) for both wavelengths.

Fig. 14.28 System composed by an anisotropic plate of unknown thickness, one polarizer and one analyzer

Solution

The only possibility for extinguishing with one polarizer any radiation, after light crosses the anisotropic material, is that the beam at the exit of the plate must be linearly polarized. For this reason, in order to ensure that the polarization of light that impinges the analyzer is linear, we will first impose the condition

$$\begin{aligned} \delta =\frac{2\pi }{\lambda }(n_{1}-n_{2})d=N\pi . \end{aligned}$$

(14.73)

Equation (

14.73 ) guaranties a linear state of polarization, but depending on

N being either even or odd, the plane of polarization may change.

To face this problem it is important to note that, what we are going to seek is the minimum thickness up to 1 mm for detecting only one of the two radiations (400 or 600 nm) that combined form the light beam. In this way, we will first use (

14.73 ) setting

\(d=1\) mm, which is the limit. This will give us the value of

N that fulfills (

14.73 ) for

\(d=1\) mm. Introducing the corresponding data we have

$$\begin{aligned} \delta =\frac{2\pi }{400\cdot 10^{-9}}(0.01)\,1\cdot 10^{-3}=2\pi \,25=N\pi \Longrightarrow N=50. \end{aligned}$$

(14.74)

Fig. 14.29 Electric fields for both lights

This result means that there is a change of the phase of light at the exit of the material. However, as

N is an even number it is equivalent to say, that no modification of the state of polarization accounts (observe that

\(\delta =2\pi .\,25\) ) if the sample would have a thickness of 1 mm. But the specimen must be thicker than

\(d=1\) mm, therefore if we substitute into (

14.73 ) the next integer number, i.e.

\(N=51\) , leaving

d as unknown, we can find out the value of

d up to 1 mm that verifies the condition required. Note that with

\(N=51\) we conserve a linear polarization state of light, which is necessary for guaranteeing the extinction of a wavelength when light crossing the analyzer for some location of it (until now unknown). However, in viewing polarization only, for

\(N=51\) it is equivalent to say that

\(\delta =\pi \) , light modifies its plane of vibration. In other words, light impinges on the sheet perpendicularly with its electric field forming an angle

\(\alpha =30^{\circ }\) with the vertical axis, and when the beam exits the material

\(\alpha =-30^{\circ }\) (with respect to

OX ). Therefore, setting

\(N=51\) into (

14.73 ) leads to

$$\begin{aligned} \delta =51\pi =\frac{2\pi }{\lambda }(n_{1}-n_{2})d\Longrightarrow d=\frac{51\pi \lambda }{2\pi (n_{1}-n_{2})}=1.02\,\,\mathrm{mm}. \end{aligned}$$

(14.75)

Now, let us make the same calculation for the radiation of

\(\lambda =600\) nm. To do this, we employ (

14.73 ) again, but for this wavelength obtaining

$$\begin{aligned} \delta =\frac{2\pi }{600\cdot 10^{-9}}(0.01)\,1\cdot 10^{-3}=2\pi \,16.7=N\pi \Longrightarrow N=33.33. \end{aligned}$$

(14.76)

This result shows that for 1 mm and

\(\lambda =600\) light changes the state of polarization, because the phase difference is not an integer number of

\(\pi \) . As far as the polarization is concerned, (

14.76 ) is equivalent to say that the beam suffers a phase change of

\(\delta =2\pi \,0.7\approx 1.33 \pi \) , which leads to an elliptically polarized light. However it does not matter with our problem. In fact, as we need that the light after crossing the plate be linear polarized, we can impose this condition by introducing

\(N=34\) into (

14.73 ), and considering

d unknown. Thus it holds

$$\begin{aligned} d=\frac{34\pi \lambda }{2\pi (n_{1}-n_{2})}=1.02\,\,\mathrm{mm}, \end{aligned}$$

(14.77)

which gives a value for

d identical that obtained for

\(\lambda =400\) . Nevertheless, as we have used

\(N=34\) (even number) in this case, neither the state of polarization of the beam nor its plane of oscillation change.

Once we know the characteristics of the light for the two radiations when passing through the anisotropic plate, we can examine how the polarizer must be located in order to avoid one of the wavelengths crossing throughout. In fact, for

\(d=1.02\) mm light comes linearly polarized in both cases, but their respective electric fields do not vibrate on the same plane. The beam of

\(\lambda =600\) nm oscillates forming an angle of

\(\alpha =-30^{\circ }\) with respect to axis

OX , then if we place the analyzer at

\(\varphi =60^{\circ }\) (see Fig.

14.28 ), this radiation will be eliminated when seeing the phenomenon after the polarizer. In this case the polarization direction and the analyzer axis are perpendicular. The light of

\(\lambda =400\) nm will be observed, but not in its totality. As the transmission axis of the analyzer forms

\(\varphi =60^{\circ }\) with

OX , the light intensity of 400 nm we will detect corresponds to those of the projection of the electric filed

\(\mathbf {E}\) over the polarization axis. Thus, the intensity of the wave leaving the analyzer will be

\(I_{p}\sim I_1\cos ^{2}30\) (see Fig.

14.29 ),

\(I_1\) being the intensity corresponding to

\(\lambda =400\) nm once light has crossed the anisotropic sheet. In the same way, we can investigate what happens with the radiation of

\(\lambda =400\) nm. This light goes out from the plate linearly polarized but its vibration axis corresponds to

\(\alpha =30^{\circ }\) , therefore if we locate the analyzer at

\(\varphi =120^{\circ }\) (see Fig.

14.30 ) we note that the electric field of this light oscillates perpendicularly to the analyzer axis. By this procedure no beam of

\(\lambda =400\) nm will be detected, but radiation of

\(\lambda =600\) nm does. In fact, for this configuration, the intensity measured is again

\(I_{p}\sim I_2\cos ^{2}30\) , where

\(I_2\) is the intensity of 600 nm observed directly behind the uniaxial plate.

Fig. 14.30 Electric fields for both lights

14.9 A sodium lamp of wavelength \(\lambda \) radiates spherical waves with circular right-hand polarization. In front of the lamp, a lens is placed in order to generate plane waves of intensity \(I_0=2\) Wm\(^{-2}\) . The beam after the lens impinges a uniaxial plate of quartz perpendicularly. The sample is cut parallel to the optic axis and its indices and thickness are \(n_1=1.5533\) and \(n_2=1.5442\) , and \(d=5.663\) mm, respectively. At the output of this plate, the light crosses a polarizer whose transmission axis is parallel to the aforementioned optic axis. (a) Obtain the polarization state of light after crossing the plate. (b) Calculate the intensity that may be registered at the exit of the polarizer. (c) If the radiation that emerges from the polarizer reaches a sheet of the figure attached under an angle of \(\theta =63.4^{\circ }\) , find the refractive index that the plate has to be in order that no reflected light beam appears (Fig. 14.31 ).

Solution

(a) When the light passes through the uniaxial plate it suffers a change in phase which may be calculated by using (

14.42 )

$$\begin{aligned} \delta =\frac{2\pi }{\lambda }(n_{1}-n_{2})d=2\pi \,87.5=2\pi N. \end{aligned}$$

(14.78)

From the point of view of the polarization, this value obtained means that the change in the phase of the perturbation is equivalent to

\(\delta =\pi \) , and therefore, the light at the exit of the anisotropic plate becomes circular left-hand polarized.

(b) The intensity of the radiation after the slab is that corresponding to circularly polarized light. If we neglects the reflections on the surface boundaries of the sample, the expression for the intensity in the most general case may be expressed as follows (see Problem 12.13)

$$\begin{aligned} I_0=\frac{1}{2}cn\epsilon _0(E^2_{x}+E^2_{y}), \end{aligned}$$

(14.79)

where

\(E_{x}\) and

\(E_{y}\) are the components of the electric field over the coordinate axes

OX and

OY , respectively. In the present problem as the light is circular polarized,

\(E_{x}=E_{y}\) , and then

$$\begin{aligned} I_0=\frac{1}{2}cn\epsilon _0(2E^2_{x})=cn\epsilon _0E^2_{x}. \end{aligned}$$

(14.80)

Now, this radiation impinges the analyzer, which only allows passing the projection of the electric field over its axis. Thus, the intensity is

$$\begin{aligned} I_p=\frac{1}{2}cn\epsilon _0E^2_{x}, \end{aligned}$$

(14.81)

and the relation between this last equation and (

14.80 ) leads to

$$\begin{aligned} \frac{I_s}{I_0}=\frac{1}{2}\Longrightarrow I_s=\frac{1}{2}I_0=1\,\mathrm{Wm}^{-2}. \end{aligned}$$

(14.82)

Fig. 14.32 Observe that the electric field of the light before impinging the slab has two components, one along OX and the other over OY . After the pass through the analyzer only the component \(E_{x}\) remains, but when light reaches the glass under conditions of Brewster’s angle it losses this component, thus no light is reflected by the surface (we only have a refracted beam at \(\theta '_{r}\) )

(c) In the former Chapter we studied the Brewster angle.

^{11} There we defined it as the angle for which the reflected light on a surface was only polarized perpendicularly to the plane of incidence (direction

OY , see Fig.

14.32 ). In the present case, because of the polarizer, we only have the component over

OX , then when reaching the glass no light is reflect but it does refracted. Applying (

14.7 ), we have

$$\begin{aligned} \tan \theta _B=\frac{n_g}{n_0}\Longrightarrow n_g=n_0\,\tan \theta _B= 1\cdot \tan (63.4)\approx 2. \end{aligned}$$

(14.83)

14.10 The system represented in the Fig. 14.33 consists of a polarizer whose transmission axis is parallel to OZ , a crystal of quartz with dielectric constants \(\epsilon _1\) and \(\epsilon _2\) , and optic axis \(\xi \) forming an angle \(\gamma \) with OX , and a screen P . A thin light beam impinges perpendicularly to the polarizer. (a) How many light points we will see on the screen P ? (b) Obtain the angle that Poynting’s vector of the extraordinary ray will form with the optic axis \(\xi \) as a function of \(\epsilon _1\) and \(\epsilon _2\) .

Fig. 14.33 System formed by a polarizer, a quarz crystal, and a screen

Solution

(a) When a plane electromagnetic radiation impinges an uniaxial slab, in principle, two independent waves orthogonally polarized to each other appear, whose phase velocities are different. In other words it means that, for any direction of the vector

\(\mathbf {k}\) we have two possible values of the wave-number

k . Taking this fact into account, for understanding what happens when a beam strikes an anisotropic plate, we can first decompose the electric field at the entrance of the material into two components, one parallel to the optic axis and the another one perpendicular to it. In the case of this problem, before the beam reaches the slab it passes through a polarizer, then at the exit of the polarizer we only have one component of the electric field, which corresponds to the projection over the

OZ axis (see Fig.

14.33 ), which in turn lies on the plane formed by

\(O\xi \) and

\(\mathbf {k}\) . As a result, only one wave propagates throughout the material, and therefore, only one point will be shown on the screen

P .

Fig. 14.34 Direction of the extraordinary ray in the material

(b) As we have seen in the theory, in anisotropic media

\(\mathbf {k}\) and

\(\mathbf {E}\) are, in general, not perpendicular to each other. This means that the direction of propagation of the wavefront

\(\mathbf {k}\) does not coincide with the propagation direction of the energy, which is represented by the Poynting vector

\(\mathbf {S}\) . For obtaining a solution to the question, we will use the principal axes of the material

\(O\xi \eta \zeta \) with

OXYZ , simultaneously (see Fig.

14.34 ). As we can observe, the Poynting vector of the extraordinary ray forms an angle

\(\beta \) with

OX , and

\(\alpha \) with

\(O\xi \) . The electric field is perpendicular to

\(\mathbf {S}\) , but not to

\(\mathbf {k}\) , therefore the angle between

\(\mathbf {E}_e\) and

OZ is the same as that between

\(\mathbf {S}\) and

\(\mathbf {k}\) . Focusing first our attention to the electric field, we can obtain a relation between its components and the angle

\(\alpha \) as follows (see Fig.

14.35 )

$$\begin{aligned} \tan \alpha =\frac{E_\xi }{E_\zeta }, \end{aligned}$$

(14.84)

where

\(E_\xi \) and

\(E_\zeta \) are the projections of

\(\mathbf {E}_e\) over the principal axes of the crystal

\(O\xi \) and

\(O\zeta \) , respectively. Along these axes it holds that

$$\begin{aligned} D_\xi =\epsilon _1E_\xi \end{aligned}$$

(14.85)

and

$$\begin{aligned} D_\zeta =\epsilon _2E_\zeta , \end{aligned}$$

(14.86)

then introduction of these values into (

14.84 ) leads to

$$\begin{aligned} \tan \alpha =\frac{E_\xi }{E_\zeta }=\frac{\epsilon _2D_\xi }{\epsilon _1D_\zeta }= \frac{n^2_2D_\xi }{n^2_1D_\zeta }. \end{aligned}$$

(14.87)

Fig. 14.35 Scheme for the vectors \(\mathbf {E}\) , \(\mathbf {D}\) , and \(\mathbf {S}\)

Now, in order to eliminate

\(D_\xi \) and

\(D_\zeta \) from (

14.86 ) we will use the diagram shown in Fig.

14.35 . This picture represents the vectors

\(\mathbf {E}_e\) ,

\(\mathbf {D}\) , and

\(\mathbf {S}\) for the extraordinary wave together with the corresponding ellipse (in two dimensions). As we can observe, it is possible to project the components of

\(\mathbf {D}\) along

\(O\xi \) and

\(O\zeta \) (

\(D_\xi \) and

\(D_\zeta \) ) over the

OZ axis of the

OXYZ coordinate frame. Then, the new components of

\(\mathbf {D}\) are

^{12} (see Fig.

14.35 )

$$\begin{aligned} D_\xi =D_z\sin \gamma , \end{aligned}$$

(14.88)

and

$$\begin{aligned} D_\zeta =D_z\cos \gamma . \end{aligned}$$

(14.89)

Substituting these equations into (

14.86 ) we obtain

$$\begin{aligned} \tan \alpha =\frac{\epsilon _2D_z\sin \gamma }{\epsilon _1D_z\cos \gamma }= \frac{\epsilon _2}{\epsilon _1}\tan \gamma =\frac{n^2_2}{n^2_1}\tan \gamma . \end{aligned}$$

(14.90)

This last equation allows us to relate the angle

\(\gamma \) formed by the optic axis with

OX (direction of

\(\mathbf {k}\) -see Fig.

14.35 ), and the angle

\(\alpha \) , but it does not give information about

\(\beta \) . However, from the Fig.

14.35 we see that

\(\alpha +\beta =\gamma \) , thus we can write

$$\begin{aligned} \tan (\gamma -\beta )=\frac{\epsilon _2}{\epsilon _2}\tan \gamma , \end{aligned}$$

(14.91)

and then

$$\begin{aligned} \beta =\gamma -\tan ^{-1}\left( \frac{\epsilon _2}{\epsilon _1}\tan \gamma \right) . \end{aligned}$$

(14.92)