Wave Optics

This project includes four independent experiments. The first three, I think, are fun and surprising. The final experiment is tedious but teaches you about useful optical devices (quarter-wave plates and half-wave plates); half-wave plates are used in the quantum entanglement experiment, for example. Experiment 1: Microwave Standing Waves

Objective:

To infer that standing waves are created by the superposition of incident and reflected waves between a microwave receiver and transmitter.

Introduction:

As students varied the distance between a microwave receiver and transmitter, they recorded the data shown in Figure 1.1.

Figure 1.1. Microwave intensity at the receiver as a function of distance between the receiver and the transmitter. (Data taken by N. Cuccia.) Why in the world should the intensity behave like this? It's certainly not a "snapshot" of the traveling wave emitted by the transmitter. How did we derive an equation to fit the data? I'll get you started. Suppose that a (sector of a) spherical wave is emitted by the transmitter. Then, the electric field along the line between the transmitter and the receiver is A E  cos(kx t) , (1.1) 1 x where A is a constant, and you probably know what everything else is. A little reflection (ha ha!) will convince you that the wave reflected off the receiver is rA E  cos[k(2L  x)  t] , (1.2) 2 2L  x where r is the amplitude reflection coefficient, and L is the distance between the transmitter and the receiver. What's the significance of (2L-x)? It's the total distance traveled by the wave. (If the reflected wave has gone halfway back, to x = L/2, it's traveled a total distance of 2L-x = 3L/2.)

Your job is to determine E3, the first wave reflected back off the transmitter toward the receiver. Then, find the superposition of E1 and E3. (If you want to add additional terms for additional reflections, have fun! I used only E1 and E3. When is this approximation valid?) If

E(L,t) = E1(L,t) + E3(L,t) (1.3) is (approximately) the electric field at the receiver, then the detected intensity is proportional to the time average of [E(L,t)]2.

Here's a suggestion: To vastly simplify the algebra, replace every cos(something) with ei(something). Then, the detected intensity will be proportional to E(L,t) × [E(L,t)]*. You don't even need to take a time average because time won't be in the result. (Why does this give the same result as the harder-math, real-number-only calculation? We can prove it if you want.)

Experiment:

Collect data, such as what's shown in Figure 1.1. There are a couple challenges. First, if the transmitter and receiver have the same polarization angle, the intensity will be way off scale. You need to adjust the relative polarization angle until it's close to 90. Second, the needle on the receiver sometimes fluctuates. I think this happens more when the intensity is low. One solution is to try to choose a range of L and a polarization angle so that the intensity is always fairly high (on the 30X scale). Another, fancier solution is to use Malus's Law to enable you to adjust the polarization as needed to keep the intensity high.

To measure distance between the transmitter and receiver, use the positions of the T marking on the base of the transmitter, and the R marking on the base of the receiver.

Curve fitting:

We can do the curve fitting by running the Python script, standing_microwaves_fit.py. Open this file in Spyder (the software that runs Python programs). First, after the word “return” in the indented line, type the expression you derived for intensity (use x for distance, and name your three fitting parameters k, R, and a). If you found that intensity simply equals 1/x, line 20 would read “return 1/x”. The symbol ** represents exponentiation, so 1/x**2 means 1/x2.

You will probably need to narrow the bounds on the fitting parameters in the line after “return.” The initial range is 0 through infinity, but we can help the program by narrowing the range. What should be the maximum allowed value of R? Can you estimate k from your data?

Instead of typing your data into Spyder, I think it’s easier if you enter it first in Excel, in two columns (distance and intensity). Run the Python script and follow the onscreen instructions to copy your data from Excel. If you don’t get a good fit, try adjusting the bounds.

Questions to answer in your lab report:

What values do you find for k and r? Are these reasonable? Does the value of r2 justify our exclusion of additional reflections? What is the microwave wavelength?

Conceptually, what causes maxima and minima in your data? Quantitatively, does your equation make sense? Look at some limiting cases. If r = 0, does the dependence of intensity on L make sense? If r = 1, does it make sense? In this case, 1/L2 is multiplied by a term that varies from 4/9 to 16/9. Why do these fractions make sense?

Experiment 2: Interference in a Glass Slide

Objective:

To determine the thickness of a glass slide by counting interference fringes.

Theory:

If you shine light on a glass slide, some will reflect off the front surface, and some will reflect off the back surface (after traveling through the slide). The interference between these two reflected beams produces dark fringes. These fringes appear as dark lines that "slide" through the reflected red spot as you vary the angle of incidence (i). You will show that the condition for destructive interference is as follows:

2dncost  m , (2.1) where d and n are the thickness and refractive index of the slide, m is any whole number,  is the wavelength of illumination, and t is the angle of transmission. Equation (2.1) can be rewritten as  cos  m . (2.2) t 2dn

Equation (2.2) says that a new fringe line appears every time cost changes by /(2dn). So if you plot cost vs. fringe number, you should obtain a line with slope -/(2dn). (The minus sign occurs because I counted fringes while increasing t; if you counted fringes while lowering t, there would be no minus sign). This is shown in Figure 2.1.

Thus if  and n are known, the slope can be used to determine d. We know that  = 632.8 nm, but n must be determined. One of the simplest ways to determine n is to find the Brewster angle (also called polarization angle): the angle at which there is zero reflection of light polarized in the plane of incidence (the plane of incidence is the plane containing the light ray and the normal to the interface). At this angle, the Fresnel coefficient rP is 0. According to the standard derivation that you can study online,

tani t  rP  . (2.3) tani t  According to Equation (2.3), rP is 0 when i + t = 90. Thus if you experimentally determine the i that minimizes reflected light, you can combine i + t = 90 with Snell's Law to find

1.000

-4 0.996 slope = -2.076 x 10 = -/(2dn)

) 0.992

t



cos( 0.988

0.984

0.980 0 20 40 60 80 Fringe number

Figure 2.1. Linear relationship between cost and fringe number.

Instructions:

 Align the glass slide so that the light is reflected directly back into the laser aperture. This is i = 0. Keeping the glass slide in this position, set the vernier scale to 0.  Now vary the angle of incidence and observe the reflected beam. You should see dark interference fringes "sliding" through the reflected red spot as you vary i.  At some arbitrary but small i, center the dark fringe in the reflected spot and record i. Call this fringe number 0. Now increase i and count fringes. Record i as a function of fringe number at least every five fringes over a minimum range of 80 fringes.  To determine n, shine the incident light through a linear polarizer so that the light incident on the slide is polarized parallel with the plane of incidence. Record i at which the reflected intensity is minimized.  Directly measure the thickness of the glass slide.

In your lab report:

 Determine n from the measured angle at which rP is minimized.  You recorded i as a function of fringe number. Compute cos(t) for each i to generate a plot like Figure 2.1. From the slope, determine d. Compare with direct measurement (with micrometer).  Derive Equation (2.1). Hints: o Referring to Figure 2.2, write AD and AC+CD in terms of d and 2.

o Next write AB in terms of d, 2, and 3. Eliminate 3 by first writing it terms of 1 and then using Snell’s law. o Next determine the optical path length difference between the two rays, remembering that the wavelength depends on the index of refraction. o Remember to think about phase change due to reflection.

1 2

1 B

A 3 D

d 2

C

Figure 2.2. Diagram used to derive Equation (2.1).

Experiment 3: Fresnel Reflection of a Light Cone

Objective:

To investigate the refraction and Fresnel reflection of a polarized light cone, through direct observation and analytical calculation.

Theory:

Figure 3.1. Schematic diagram of the experiment. The small arrows sticking out of the light rays represent polarization (vertical, horizontal, or both).

Figure 3.1 represents the major elements of the apparatus. A uniform, collimated beam approaches a polarizing beamsplitter cube from the left. The original beam may be unpolarized. The polarizing beamsplitter cube transmits the horizontally polarized component, while reflecting the vertically polarized light into a beam block (or just into the room to be ignored). The light approaching the microscope objective lens is uniformly horizontally polarized, as shown in Figure 3.2. The microscope objective focuses the light onto the glass slide. Some of the light reflects, and some is transmitted, according to the Fresnel coefficients rp and rS. Please look these up online or in a textbook.

The light that reflects off the glass slide returns through the objective lens, becoming a collimated beam returning toward the polarizing beamsplitter. The polarizing beamsplitter cube again transmits the horizontally polarized component of each ray while reflecting the vertically polarized component onto a screen. You will see a faint but unmistakable “Swiss cross” pattern on the screen. Where did it come from? Why is there any vertically polarized light at all in the beam reflecting from the glass slide? Recall that all the light approaching the glass slide was horizontally polarized (Figure 3.2). y polarization vector at this point (along x axis) is in the radial direction

x

polarization vector at this point (along y axis) is in the tangential direction

Figure 3.2. The polarization vectors of a circular beam of light. The horizontal axis is x, and the vertical axis is y.

To solve this mystery, we need to think about how to use the Fresnel coefficients for each ray in Figure

3.2. The Fresnel coefficient rP applies to light polarized in the plane of incidence (the plane that includes the light ray and the normal to the interface). The Fresnel coefficient rS applies to light polarized perpendicular to the plane of incidence.

First consider a light ray along the x axis in Figure 3.2. The polarization is in the radial direction. When we follow the light ray through the microscope objective to the glass slide in Figure 3.3, we see that the electric field of light polarized in the radial direction is in the plane of incidence. When we examine a light ray on the y axis, we see that it is polarized in the tangential direction, and the electric field of light polarized in the tangential direction is perpendicular to the plane of incidence.

What about an arbitrary light ray in the beam, located on neither x nor y axis? The electric field of this ray must be decomposed into radial and tangential components, as shown in Figure 3.4. In this figure,  and  are radial coordinates, i and t are angles of incidence and refraction, h is the focal length of the microscope objective, E0 is the amplitude of the electric field, Ei is the electric field of the incident ray, Ef is the electric field of the reflected ray, and  is the half-angle of the light cone.

The light reflects off the glass plate according to the Fresnel coefficients; the perpendicular and parallel components are reflected in different proportions. Thus, the polarization vectors of the reflected beam will not be in the same direction as those of the incident beam. The resulting polarization vectors are shown in Figure 3.5 for ni = 1 (air’s refractive index), nt = 1.5 (the assumed refractive index of the glass slide, and NA = 0.9. (NA is the numerical aperture of a microscope objective. NA defined as sin. You can write it in terms of the beam radius and the focal length of the microscope objective.)

y

x

polarization polarization vector vector in radial in tangential direction is direction is parallel with perpendicular to plane of plane of incidence incidence (xz (yz plane) plane)

Figure 3.3. The electric field of the ray on the x axis is parallel with the plane of incidence (xz), and the electric field of the ray on the y axis is perpendicular to the plane of incidence (yz plane).  ˆ Ef  rPE0 cosˆ rS E0 sin

y ˆ D ˆ rPE0cos r E sin  S 0 x E0cos A  ˆ Ei E0i E0sin

E0cos

rPE0cos B C E0sin h rSE0sin

i



t

Figure 3.4. Fresnel reflection of an arbitrary light ray.

Figure 3.5. Polarization vectors for the reflected beam (before it reaches the polarizing beamsplitter cube).

Figure 3.5 illustrates that the polarization vectors are, in effect, rotated, depending on their positions in the beam. In contrast with Figure 3.2, the reflected beam is not uniformly polarized, and it is not entirely horizontally polarized. You can see from Figure 3.5 that the four "corners" of the beam contain the light with the greatest y components of polarization. Thus, retaining along the y polarization, the light observed will appear as shown in Figure 3.6.

Figure 3.6. Final light intensity observed on the screen.

Instructions:

 Laboratory experiment: o Confirm that the polarizing beamsplitter cube transmits horizontally polarized light and reflects vertically polarized light. (This is true for the chosen orientation of the cube; more generally, it transmits light with polarization parallel to the plane of incidence at the diagonal surface, and it reflects light with polarization perpendicular to the plane of incidence at the diagonal surface.) o Design and build a beam expander to broaden the laser beam (but keep it collimated!). o Shine the expanded beam through the polarizing beamsplitter cube. o Next, focus the beam onto a glass slide using a microscope objective with NA = 0.9. Observe the reflected pattern.  Analytical calculation: Compute intensity as a function of x and y to generate a figure like Figure 3.6. Hints: o Figure 3.5 shows the electric field Ef of the reflected beam before it is further reflected by the polarizing beamsplitter cube. The polarizing beamsplitter cube reflects the y ˆ component of each ray, E f  j . Light intensity (I), which is what we observe, is  ˆ 2 proportional to the square of the electric field, so I  E f  j . This is a function of position (x,y) or (,) in the beam. o To evalulate the dot product, consider: ˆj

ˆ

ˆ

 iˆ

o To organize your thoughts about the sequence of calculations to perform for each ray in the beam (each “pixel” in Figure 3.6), consider:

x y   i t rP rS I

o Since intensity I is a matrix, each parameter above should be a matrix. The x matrix will consist of identical rows, and the y matrix will consist of identical column. o You may arbitrarily set beam radius to 1, but NA should be 0.9 since that’s the NA of our microscope objective.

Experiment 4: Retarders

Objective:

To study the effects of quarter-wave and half-wave plates on linearly polarized light.

Theory:

Linear polarization occurs when the electric field of a light wave travels in a single plane; at any position, the electric field vector grows, shrinks, and reverses direction along a single line. Circular polarization occurs when the electric field vector at any position rotates in a circle. Both linear and circular polarization might be considered special cases of elliptical polarization, in which the electric field vector at any position traces out an ellipse.

To study polarization mathematically, consider an electromagnetic wave traveling in the z direction. The electric field vector may be written generally as  ˆ ˆ E  iEx cos(kzt)  jEy cos(kzt  ) . (4.1)

If ∆ = 0 (or any integral multiple of ), the light is linearly polarized in a direction determined by the superposition of the x and y components. If Ex = Ey and ∆ = /2 (or any odd integral multiple thereof), the light is circularly polarized.

We begin by shining light through a linear polarizer whose transmission axis is rotated an angle  counterclockwise from the x axis (which I take to be the downward direction because the polarizer mount indicates angles relative to the downward direction). The electric field of the resulting linearly polarized wave is  ˆ ˆ E  iE0 cos cos(kzt)  jE0 sin  cos(kzt) . (4.2)

This linearly polarized light is then passed through a retarder. A retarder is a device that has a fast axis and a slow axis. Let's align the slow axis with the x axis. The retarder will then cause the x component to be shifted ∆ in phase relative to the y component:  ˆ ˆ E  iE0 cos cos(kzt  )  jE0 sin  cos(kzt) . (4.3)

Confirm that if ∆ = /2 and  = 45, the light will be circularly polarized; if ∆ =  (for any ), the light will still be linearly polarized, but with the plane of polarization flipped about the y axis.

Finally, the light is passed through another linear polarizer at an angle  relative to the x axis. The magnitude of the transmitted wave is found by projecting Equation (3) onto the axis of transmission:

E  E0 cos cos(kzt  )cos  E0 sin  cos(kzt)sin  . (4.4)

We know that the measured light intensity is proportional to the time average of E2. You're invited to derive the result:

1 1 1 I ~ E 2 cos 2  cos 2   E 2 sin 2  sin 2   E 2 sin( 2)sin( 2 )cos() . (4.5) 2 0 2 0 4 0

There are two special values of ∆: /2 (quarter-wave plate, so called because /2 is one quarter of a full 2 phase shift) and  (half-wave plate). For a quarter-wave plate, cos(∆) = 0 and the last term drops out of Equation (4.5). If, furthermore,  = 45 (circular polarization), I becomes independent of , as you should confirm. This is because a constant fraction of circularly polarized light is transmitted through a linear polarizer, regardless of the orientation of the polarizer.

For a half-wave plate, Equation (4.5) simplifies to 1 I ~ E 2 cos2 (   ) . (4.6) 2 0

Instructions:

 Let a HeNe laser warm up for about an hour. I’ve noticed that the polarization of the laser is unstable at first. You may investigate this if you like.  Shine the laser light through the following sequence of devices: a linear polarizer at angle , a 140 nm retarder, and a second linear polarizer at angle . Align the slow axis (the 0-180 line) of the retarder with the x direction and set  = 45. o Observe on a screen that the light intensity passing through the second polarizer appears nearly independent of . o However, the polarization is not perfectly circular because ∆ = 2  140/650 = 1.35 instead of /2 = 1.57. Using the photometer, observe the dependence of the light intensity on . Record the light intensity as a function of , varied in 10 increments. Compare the resulting curve with Equation (4.5). If you don’t do a curve fit, at least compare the positions of maxima and minima and on the flatness of the curves.  Shine the light through the following sequence of devices: a linear polarizer at angle , a half-wave plate (two quarter-wave plates in a row), a second linear polarizer at angle , and the photometer. Align the slow axis of the retarder with the x direction. o Choose five settings for ; for each, find the  that maximizes the transmitted power. Compare your findings with the  values you'd predict from Equation (4.6). o For one value of , observe the dependence of the transmitted light intensity on . Record the light intensity as a function of , varied in 10 increments. Compare the resulting curve with Equation (4.6). What is the polarization of the light emerging from the half-wave plate? o If you have time for a more detailed analysis, use Eq. (4.5) with ∆ = 2  140/632.8 = 1.39.