Lecture 15: Fabry-Perot Spectroscopy,
worked examples, comparison to diffraction
by many slits
Lecture aims to explain:
1. Fabry-Perot spectroscopy, examples
2. Spectral resolution using Fabry-Perot interferometer
3. Free spectral range of Fabry-Perot interferometers
and assemblies using many slits (grooves)
Fabry-Perot Interferometer, reminder
glass
etalon
focal plane
Normalised transmitted intensity
ground
1.0
lens
0.8
0.6
0.4
0.2
0.0
2
4
6
8
10
12
14
Phase difference δ
δ=
4πnt cos θ
λ
16
Calculate the intensity (relative to
maximum) of the transmitted light exactly
in the middle between the adjacent
fringes for etalons with F=1, 10 and 100.
Calculate FWHM (in terms of phase) of
the fringes in these etalons.
1.0
Normalised transmitted intensity
Example 15.1
0.8
0.6
F=100
F=10
F=1
F=0.2
0.4
0.2
0.0
2
4
6
8
10
12
Phase difference δ
14
16
Raleigh criterion for Fabry-Perot spectroscopy
γ
Fringe intensity
1.0
Spectral resolution criterion: two close wavelengths
are just resolved if the cross-over of the two fringes
in interference patterns of these two wavelengths is
at the half maximum of intensity
λ1
λ
0.5
The resolving power of a Fabry-Perot
interferometer
0.0
δ0 δ1
Phase δ
λ
mπ F
mπR
=
=
∆λ
2
( 1-R 2 )
Example 15.2
Light with wavelength 500 nm passes at normal incidence through the
interferometer made from a 0.1 cm thick glass plane parallel plate (n=1.5)
coated with silver. By gradually titling the plate the periodic change in
intensity is observed from I0 to 0.01 I0.
(i) Explain the origin of this observation
(ii) Calculate the coefficient of finesse of this etalon.
(iii) Calculate the resolving power of this etalon.
(iv) Calculate the angle corresponding to the 10th minimum. Explain how
this can be used to measure the wavelength.
(v) By how much the wavelength should be tuned so that a maximum is
observed at the angle corresponding to the 10th minimum.
Example 15.3
A Fabry-Perot interferometer operating at normal incidence is made of two
mirrors with R=0.9 with a 1 mm air gap between them. Calculate the
resolving power of this instrument in the wavelength range centred at 500
and 1000 nm.
Free spectral range
We need to consider the problem of overlapping orders
leading to the notion of the free spectral range.
The maxima of two wavelengths overlap when
m( λ0 + ∆λ ) = ( m + 1 )λ0 ⇒ ∆λ = λ / m
The free spectral range is given by
( ∆λ ) fsr = λ / m
m is a few 1000 for a Fabry-Perot interferometer leading to an extremely
small free spectral range
Example 15.4: assembly with many slits
Light is diffracted by a 5 cm x 5 cm screen having 100 slits/mm. Light
is then focussed by a lens on a detector array. Find an optimum size
of such spectrograph with a maximum possible free spectral range if
a typical size of the detector pixel is 20 microns in size.
SUMMARY
γ
The resolving power of a Fabry-Perot
interferometer
λ
mπ F
mπR
=
=
∆λ
( 1-R 2 )
2
This corresponds to the cross-over of the
two fringes in interference patterns of the
two close wavelengths at the half maximum
of intensity
Fringe intensity
1.0
λ1
λ
0.5
0.0
The free spectral range corresponds to the range of
wavelength which could be measured without overlap.
The FSR is given by:
( ∆λ ) fsr = λ / m
δ0 δ1
Phase δ