PHYS 320, Spring 2013
8.1)
HW 8 Solutions
Due: April 5, 2013
Pedrotti3 8-1
 2d   2  0.014 cm 
 4.6  105 cm 

523
 m 
 
8.2)
Pedrotti3 8-4
8.3)
Pedrotti3 8-6
436 nm
PHYS 320, Spring 2013
8.4)
Pedrotti3 8-8
8.5)
Pedrotti3 8-10
HW 8 Solutions
Due: April 5, 2013
PHYS 320, Spring 2013
8.6)
Pedrotti3 8-14
HW 8 Solutions
Due: April 5, 2013
PHYS 320, Spring 2013
8.7)
Pedrotti3 8-19
HW 8 Solutions
Due: April 5, 2013
PHYS 320, Spring 2013
8.8)
HW 8 Solutions
Due: April 5, 2013
Pedrotti3 8-21 Consider a light source consisting of two components with different
wavelength 1 and 2. Let light from this source be incident on a scanning Fabry-Perot
interferometer on nominal length d = 5 cm. Let the scaled transmittance through the length be
as shown in Figure 8-20a and 8-20b. Figure 8-20b shows the first set of dual peaks of Figure
8-20a over a smaller length scale in order to allow a closer examination of the structure of the
overlapping peaks.
a.
b.
c.
What is the nominal wavelength of the light source?
Estimate the difference 2  1 in wavelength of the two components presuming that the
overlapping transmittance peaks have the same mode number, m2 = m1 = m.
Estimate the difference 2  1 in wavelength of the two components presuming that the
overlapping transmittance peaks have mode numbers that differ by 1, so that m2 = m1 +1
 0.002 μm 
Focus: Use relationship between the FabryPerot cavity length FSR and the cavity
frequency FSR (and the frequency relationship
to the wavelength)
   0.50 μm 
Plan:
Part c) m2 = m1 +1
2d 2d
2  1  2  1
m2 m1
2m d  2m2 d1
2  1  1 2
m1m2
d fsr 

2
2d
n  n
mn
  2.0 108 μm
Execute:
Part a)

d fsr 
2
  2d fsr
2  1 
From Figure 8-20, d fsr  0.25 μm
2  1 
  0.50 μm
Part b) m2 = m1 = m
2d 2d
2  1  2  1
m
m
2d
2  1 
2d1 1
 d

1
d1
d
  1
d1
50000 μm
2  1 
2m1d 2  2  m1  1 d1
m1  m1  1
2m1d 2  2  m1  1 d1
m1  m1  1
2m1  d1  d   2  m1  1 d1
m1  m1  1
2m1d  2d1
m12
Now, m1  2d1   2d 
2  1 
2  1  
d  2

d 2d
2  1   2.0  10
 0.50 μm 
μm  
2  50000 μm 
2
8
2  1  2.48 106 μm
Evaluate: Units are correct.
PHYS 320, Spring 2013
8.9)
HW 8 Solutions
Due: April 5, 2013
Explain the difference between wavefront division and amplitude division in an
interferometer. Give an example of each.

Amplitude division: The E-field is separated b y a beam splitter such that part
of the energy is directed in two directions (ex. Michelson Interferometer).

Wavefront division: The wavefront is separated spatially on two locations of
the wavefront, usually at an aperture stop, and those parts are superimposed to
produce fringes (ex. Young’s Double Slit)
8.10)
What is the difference between fringes of equal thickness and equal inclination?

Equal thickness fringes – contour maps of separation between two surfaces
that have an equal separation distance.

Equal inclination (Haidinger) fringes are a function of the angle of inclination
on surfaces formed by parallel beams from a source. Fringes are formed by
rays reflected at a given angle.