FP Phase Differences and Camera Requirements
Path Length Differences for parallel ray bundles passed at different angles through a FP etalon.
Review of the relation between wave phase and rays and wavefronts. The discussion is restricted to propagation in isotropic media.
Wavefronts are surfaces of constant phase, and rays are point in the direction of the most rapid change of phase. That is: rays and wavefronts are locally perpendicular.
rays and wavefronts
In the case of parallel rays, equi-phase wavefronts are planes perpendicular to the rays.
FP Phase.1

FP Phase Differences and Camera Requirements
In the case of interfering rays, the interference is constructive if the phase difference is an integer number of cycles. This difference often arises from a difference in path lengths traveled. In the absence of other phase changes, the condition for constructive interference is that the difference in path lengths is an integer multiple of the wavelength of the light in the medium . In addition, there is a phase change of

when a ray traveling in one medium is reflected at an interface with a material of higher index of refraction. There is no phase change for reflection from a lower index or upon transmission through the interface.
An air spaced etalon is two very flat parallel surfaces each with reflectivity R . The focusing lens brings a family of parallel rays to a point on its focal plane at the point that the un-deflected ray through the lens center crosses the focal plane. (The substrates on which the mirrors coatings reside are wedged to ensure that the reflections from the opposite surfaces do not interfere.) The hall of mirrors behavior leads to an effective number of interfering beams equal to F =

1

R
R , the reflecting finesse.

The correct operation of the etalon requires that the multiply reflected rays remain parallel. The condition that this be true is that the mirrored surfaces are absolutely parallel. (We will need to trim the mirror alignment to get as near to this condition as possible.)
The Condition for Constructive Interference :
FP Phase.2

FP Phase Differences and Camera Requirements
A direct ray is one entering from the left that is transmitted through the first mirror that then propagates between the mirrors and is transmitted through the second mirror. Reflected ray 1 is reflected at the second mirror and travels back and forth between the mirrors being transmitted on its second encounter with the right side mirror.
Find the additional path length traveled by reflected ray 1 as compared to the direct ray. Recall that the equi-phase wavefronts are perpendicular to the rays. d cos q
2q d cos q q d d cos q d cos q
2q q d
Path Difference 1 Path Difference 2 reflect 1 direct
The direct ray and reflected ray 1 separate at the red dot. The mirror separation is d , and the rays propagate at an angle q
relative to the surface normal. A ray travels d / cos q to return to the left mirror and cos(2 q
) d / cos q to travel from the left mirror to
FP Phase.3

FP Phase Differences and Camera Requirements reach the wavefront (surface of constant phase) to compare phase with respect to the direct ray wavefront through the red point. Summing the contributions, the net path length difference for the two rays is 2 d cos q
. The condition for constructive interference for q
= 0, the spot in the center of the ring pattern, is that 2 d = m
 where m is the order of the interference. For d = 2 mm and

= 500 nm , m is about 8000. For larger q
, 2 d cos q
= m
 
(which is less than 2 d ) so the orders of the for the rings are m - 1, m - 2, … as the rings increase in radius. The shorter path difference for increasing q
means that to maintain constructive interference on m th order as you move out, the light must increase in frequency (as

decreases).
m

= 2 d; m

= 2 d cos q
same order m

decreases as q
increases

Exercise : Show that: d q
 d cos cos q q 
2 cos q
.
Exercise : Show that the path length difference between the direct ray and the reflected ray 1 measured from the red dot where they split to the common wavefront after the etalon is 2 d cos q
. See figure Path Length Difference 2.

Estimate for Required Lens Focal Length
For our case, d = 2 mm so the interference order for the center spot is (4 mm)/

or about max = 7100. We might wish to see 3 or 4 orders.
7100 – 4 = (2 d /

) cos q max
= 7100 – 4 = (2 d /

) [1- ½ q 2 max
]
4/7100 = ½ q
2 max
or q max =
0.03356 = 1.9
o
The sensor is 1 /
3
”
or 8.47 mm format. Using 4.2 mm for one side the center spot plus 4 rings requires f = 4.2
/
0.03356
= 126 mm or for only half rings f = 63 mm. The target focal length for the lens is to be 100 mm.
FP Phase.4

FP Phase Differences and Camera Requirements
100 mm for 179$ look for 1/3 format
The free spectral range of an air-spaced etalon is the frequency increase given constructive interference at q
= 0 that would lead to constructive interference at q
=
0 for order of interference m
+
= m + 1. m
 
( m

1)


, m c f

( m

1) c
( f
 f )

( m

1) c f
(1
  f f
)

FP Phase.5

FP Phase Differences and Camera Requirements c f

( m

1) c f
(
 f f
) or f m
Note that m is large compared to 1. For the original frequency, m

= 2 d or f =
 m ( c /
2 d
). Hence
 f = ( c /
2 d
) or 75 GHz for an etalon with air spacing of 2 mm . The quantity
 f is the free spectral range which will be discussed below.

Note that |
 f / f
| is about |

/ 

so a free spectral range of 75 GHz corresponds to

of about 0.0625 nm for light with a wavelength around 500 nm .
Exercise : Given that f = c /  , show that
 f / f
is about -

/  . Find the frequency for light with wavelength 546.07 nm.
Find

for a frequency increase of 75 GHz.

What is the finesse? The finesse is approximately the number of individual lines that can the etalon can resolve in a free spectral range.
FP Phase.6

FP Phase Differences and Camera Requirements
The spectrum overlays and repeats each free spectral range. For the example above, the free spectral range is
 f fsr
= 25 GHz 1 and the resolved linewidth for a single line is the
 f res
or about 500 MHz. The finess is therefore 25 GHz /
500 MHz
= 50.
The finesse for our Zeeman experiment is less than 20. The importance of the free spectral range can be seen when one views the mercury ring pattern without any filters. The yellow, green, violet and deep violet rings are all visible at the same time. Hence the spectrum is cluttered, not free. The same wavelength appears at multiple positions. The violet dominates, but you can view through the green line filter and observe that a little yellow (577/579 nm ) leaks through and clutters the spectral display.
Free spectral range and spectrum overlay: An etalon spearates the spectrum from from f o
to f o
+
 f fsr
into a one to one display. The spectrum from f o
+
 f fsr
to f o
+ 2
 f fsr
is overlaid on the f o
to f o
+
 f fsr
range. Next the f o
+ 2
 f fsr
to f o
+ 3
 f fsr range is overlaid on the first two and so on. In order to remove the confusion, the etalon must be used in tandem with a pre-selection (or post-) device which effectively restricts the input to the etalon to one free spectral range. This task is fairly simple as quantum mechanics limits the visible radiation from neutral mercury to a few narrow spectral lines with separations of several nm . Band pass filters can easily restrict the transmitted light to a 10 nm band which allows us to single out the 546 nm and the 436 nm.
(We might be able to single out the 405 nm and the 577/579 nm doublet with additional filters.) In the case of the green line pass filter, we see that its poor design does permit some transmission at 577 nm
(yellow) as well as around the intended 546.1 nm (green) .
Our tandem system is a band pass filter for pre-selection followed by the etalon.
1 Our etalon has a fsr of 75 GHz.
FP Phase.7

FP Phase Differences and Camera Requirements
The focusing lens:
Cente r ray f
The lens is designed so that the light in parallel beams that have equal phase as they cross the orange wavefront before the lens will arrive with the equal phases at the image point a focal distance after the plane of the lens. That point is where the center ray crosses the focal plane. The lens is designed so that each of the rays has the same time to travel from the pre-lens wavefront to the common spot of the focal plane. Rays father front he center ray travel farther, but through less glass, which delays the light. Hence they propagate to the focal plane spot in the same time as the center ray. As the rays start in phase and arrive after the same delay, they interfere constructively at the focal plane image spot. By extension, not all of the rays in this bundle interfere constructively at any other point in the focal plane.

FP Phase.8

FP Phase Differences and Camera Requirements
In the thin lens approximation each ray path from the source point to the image point requires the same time so that rays leaving the source arrive in phase at the image point.
FP Phase.9