5. A Michelson interferometer uses a sodium lamp as a light source. Assume that

Question

5. A Michelson interferometer uses a sodium lamp as a light source. Assume that the sodium lamp is monochromatic and has a wavelenth of 589.0 nm.

a) How far would we have to move the translating mirror to see 1000 fringes pass some point in the pattern?

b) If the optical path length difference between the two arms is 2.0000 mm, what is the angular diameter of the second bright fringe from the center of the pattern?

c) If we insert a piece of glass (thickness = 0.200 mm, n=1.45) in one arm of the interferometer how many fringes would pass the center of the pattern, assuming that the glass is wedged so that the thickness increases from 0 to 0.200 mm as the glass is inserted?

Explanation / Answer

a)
given,

wavelength lambda=589nm=589*10^-9 m

the distance moves by mirror is,
(x2-x1)=N*lambda/2
=1000*(589*10^-9)/2
=294.5*10^-6 m
=0.2945 mm...is answer

b)
for the central bright fringe optical path difference is
2d*cos(angle)=n*lambda
for the central bright fringe the angle is =0degrees
now
2d*cos(0)=n*lambda
2d=n*lambda .......(1)

for the second bright fringe optical path difference is,
2d*cos(angle)=(n-1)*lambda ....(2)

now,
by solving (1) and (2)

from (2)-----------> 2d*cos(angle)=(n-1)*lambda

2d*cos(angle)=n*lambda-lambda

but from (1) 2d=n*lambda

hence
2d*cos(angle)=2d-lambda
cos(angle)=(2d-lambda)/2d
cos(angle)=1-lambda/2d
cos(angle)=1-589*10^-9/2*2*10^-3
=====> angele=0.98 degrees
angular diameter=0.98 degrees ...........is answer
or
angular diameter=1degrees(approximate)...is answer

c)
if we insert a glass of refractive index n and thickness t,

path difference is,2(n-1)*t=N*lambda

from tha above equation
no. of the fringes is N=2(n-1)*t/lambda
N=2(1.45-1)*(0.2*10^-3)/589*10^-9
N=305.6 ..is answer
or
N=306 .....

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