The Fabry-Perot Interferometer: Making Use of Optical Cavities

Transcription

1 The Fabry-Perot Interferometer: Making Use of Optical Cavities One type of Fabry-Perot device consists of a pair of flat, parallel, partially transmitting mirrors. If the distance between the mirrors is fixed the device is called an etalon, if variable it is called an interferometer. Although we should employ a wave description when computing the action of this device, a simpler method which gives the basic properties is to use a ray description and sum the multiple reflections from the two mirrors. For simplicity, we will assume identical mirrors. We take t = amplitude transmission coefficient for the mirrors and r = amplitude reflection coefficient. A wave with complex wave amplitude E 0 incident on a single mirror produces a transmitted wave of complex amplitude E 0 t and a reflected wave of complex amplitude E 0 r. Problem: Show that for a single mirror the ratio of reflected to incident intensity is r 2 while the ratio of transmitted to incident intensity is t 2. Deduce that in the absence of absorption r 2 + t 2 = 1. Multiple reflections produce two types of waves, those that are ultimately reflected back toward the direction of the incident light and those that are transmitted forwards. Let be the angle between the incident light and the normal to the plane of the mirror. We need to find the phase difference between two successive transmitted waves. The situation is similar to that discussed in section 8 and equation (8.5) can be applied directly. Each round trip between the mirrors advances the the complex amplitude by a factor e i where = 2 k d cos. Defining ( 1 = E 0 e i /2, the transmitted wave has complex amplitude ( T ( 1 t 2 +E 1 e i t 2 r 2 + E 1 e 2i t 2 r = ( 1 t 2 ( 1 + e i r 2 + e 2i r ) i.e. (12.5) ( T = ( 1 t 2 / ( 1 - e i r 2 ) so that the average intensity is (12.6) I T = I 0 t 4 / 1 - e i r 2 2 Here I 0 is the intensity of the incident light. The intensity reflection coefficient is R = r 2 and we can write r = R e i where is the phase of the wave reflected from a single mirror relative to the 17.1

2 phase of the incident wave. Similarly, the intensity transmission coefficient is T = t 2. We therefore have (12.7) I T = I 0 T 2 / 1 - R e i 2 where = + 2. Now, (12.8) 1 - R e i 2 = 1 + R 2-2R cos and cos = cos ( /2 + /2) = cos 2 ( /2) - sin 2 ( /2) = 1-2 sin 2 ( /2), so that (12.9) 1 - R e i 2 = (1 - R) 2 + 4R sin 2 ( /2) (1 - R) 2 [1 + F sin 2 ( /2) ] where F = 4R/(1 - R) 2 is called the finesse. It follows that (12.10) I T = I 0 T 2 (1 - R) -2 [1 + F sin 2 ( /2) ] -1 The first factor in (12.10) is determined by the incident light, the second and third factors by the properties of the individual mirrors and the last factor by the interference which depends on the wavelength, angle of incidence, and mirror separation. The factor [1 + F sin 2 ( /2) ] -1, regarded as a function of, is called the Airy function. Problem: A. What are the positions (i.e. the values of ) of the intensity maxima predicted by (12.10)? What values of result in the phase difference between successive transmitted waves being 2? B. Plot the Airy function for various values of R. Comment on the implications of your plots for the fringe visibility. C. R is fixed by the properties of the mirrors but can be controlled by the experimenter (how?). Show that with R fixed and variable the fringe visibility is equal to F/(2+F). D. The illustration on the next page shows a typical peak as encountered in many physics experiments. A criterion for the width of the peak is the so-called full width at half maximum (FWHM). Use (12.10) to show that for the peaks plotted in part A above, whenever it is legitimate to replace sin by the FWHM is = 4 /F. 17.2

3 Resolving Power of the Fabry-Perot Interferometer The Fabry-Perot interferometer s ability to distinguish nearly equal wavelengths is unsurpassed. To see why this is so, consider two wavelengths 1 and 2 with = 1-2 and suppose that << 1 or 2. Each spectral component produces interference fringes as considered above and the two fringe systems will nearly overlap. Let us take as a criterion that two fringes will be considered distinguished if the cross-over point between the two intensity curves occurs at the half-maximum position. This is called Taylor s criterion. (Why can t we use Rayleigh s criterion?) 17.3

4 The separation between the two peaks is 2-1 = = 4 /F as shown in the previous problem. If we consider the typical case where = we can take = = 4 d cos./ so that when the two spectral components satisfy Taylor s condition we have (12.11) 4 d cos./ 2-4 d cos./ 1 = 4 / F If is the mean of 1 and 2 then (12.11) can be rearranged to give (12.12) = 2 / [ d cos F ] As an example, if r = 0.95, = 500 nm, and d = 1 cm, we have F = 380 and = nm. The resolving power of an interferometer is defined as (12.13) RP / so our hypothetical instrument has RP 10 6, meaning that it can resolve wavelengths to a part per million. For comparison, a typical grating spectrometer has RP

5 Experiment 12 : Finding the Wavelength of a Laser with a Fabry-Perot Interferometer Apparatus: Pasco precision interferometer set up in Fabry-Perot mode, HeNe laser Procedure: 1. Align the laser and base so that the laser beam strikes the center of the moveable mirror and is reflected directly back into the laser aperture. 2. Mount the adjustable mirror where indicated on the interferometer base and place one component holder in front of the adjustable mirror and another behind the moveable mirror. Attach the viewing screen to its magnetic backing. 3. You will see several images of the laser beam on the viewing screen. Using the thumbscrews, adjust the tilt of the sdjustable mirror until there is only one bright dot on the screen. 4. Mount the f = 18mm lens on the front component holder. A clear, sharp interference pattern will be visible on the viewing screen if the apparatus has been set up properly. The fringes are extremely sensitive to slight changes in the orientation of the adjustable mirror. As usual with interferometry, patience is required. 5. The experiment is simplest if the viewing screen is removed and the fringes are projected onto a screen a few meters away. Pick a fixed point on the observing area (such as the outer rim of the illuminated region), note the micrometer reading and count at least 100 fringes. Each experimenter should do this at least twice. Average your four values and compute. 17.5