Haidinger fringe (interferometer at zero path difference)

In summary: However, through my calculations, it seems like the mirror would only need to be moved a very small distance in order to set the interferometer at zero path difference. This could potentially make sense since the laser used has a relatively short wavelength and the fringes in the image are relatively small in diameter. So in summary, the mirror that is further away from the beamsplitter would only need to be moved a very small distance, likely less than 1 cm, in order to set the interferometer at zero path difference.
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Homework Statement


A Michelson interferometer is illuminated with a laser with a wavelength of 514.5nm. A Haidinger fringe pattern is photographed with a lens of focal length 55mm. The diameter of the two adjacent circular fringes in the image are 1.53mm and 2.62mm.

How far would the mirror that is further away from the beamsplitter need to be moved in order to set the interferometer at zero path difference?

Homework Equations


Haidinger Fringe rp
rp=f*√[(p*λ)/d]

where
rp = Haidinger Fringe
f = focal length
p = order of interference
λ = wavelength
d = difference in distance between the two mirrors and beamsplitter

Effective path difference
2d*cosθ = pλ

The Attempt at a Solution


Trying to derive an equation for the two path differences from the Haidinger Fringe equation, which is independent of the p value, but I'm struggling...

rp=f*√[(p*λ)/d]

( rp / f )2 = (p*λ)/d

( rp / f )2 = 2d*cosθ/d

d = 2d*cosθ / ( rp / f )2

d = 2d*cos1.3644 / 1.31 mm/55mm

d = 2*cos1.3644 / 0.02381818181

d = 17.2082026858

2 * 17.2082026858 * cos1.3644 = pλ

p = 2 * 17.2082026858 * cos1.3644 / 5.145*10^-7

p = 7.05309334257 / 5.145*10^-7

p = 13708636.2344

5.145*10^-7 * 13708636.2344 = 7.05309334257 metres = difference in distance between the two mirrors and the beam splitter

So would the mirror need to be moved 7.05309334257 metres in order to set the interferometer at zero path difference?
 
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I made a new attempt. Bare with me!

Homework Statement


A Michelson interferometer is illuminated with a laser with a wavelength of 514.5nm. A Haidinger fringe pattern is photographed with a lens of focal length 55mm. The diameter of the two adjacent circular fringes in the image are 1.53mm and 2.62mm.

How far would the mirror that is further away from the beamsplitter need to be moved in order to set the interferometer at zero path difference?

Homework Equations


says said:
Haidinger Fringe rp
rp=f*√[(p*λ)/d]

where
rp = Haidinger Fringe
f = focal length
p = order of interference
λ = wavelength
d = difference in distance between the two mirrors and beamsplitter

Effective path difference
2d*cosθ = pλ
[/B]

The Attempt at a Solution

[/B]
Haidinger Fringe, rp

rp = f √ (pλ/d)

rp = f [ (( 1 - pλ/d )^-2) - 1 ] ^(1/2)

cosθp = f / √ ( rp2 + f2 = [ 1 - (pλ / 2d) ]

2d(1-cosθp) = pλ

∴ rp = f [ (( 1 - 2d(1-cosθp/2d )^-2) - 1 ] ^(1/2)

A lot of math later and I get a value of...

p = 55
d = 5cm

∴ Effective path difference
2d*cosθ

2*0.05*0.999716484 = 0.0999716484 m

0.999716484 cm

The mirror would have to be moved 0.999716484 cm in order to set the interferometer at zero path difference.

I'm not sure if this is correct. I've not had much experience with interferometers or the theory behind them.
 
Last edited:

What is Haidinger fringe?

Haidinger fringe is a visual phenomenon where faint yellow or blue bands can be seen at the center of a black cross or circle. It is caused by the interference of light waves when they pass through the human eye's lens and cornea.

How is Haidinger fringe related to interferometers at zero path difference?

Haidinger fringe is often observed in interferometers at zero path difference, which are devices used to measure the properties of light waves. The presence of Haidinger fringe indicates that the light waves are interfering with each other, which is a key principle of interferometers.

What causes the yellow and blue colors in Haidinger fringe?

The yellow and blue colors in Haidinger fringe are caused by the interference of two polarized light waves. When these waves overlap, they create a pattern of alternating bright and dark fringes, with the yellow and blue colors representing the highest and lowest points of interference, respectively.

Can Haidinger fringe be observed in all interferometers at zero path difference?

No, Haidinger fringe can only be observed in interferometers that use polarized light. Interferometers that use unpolarized light will not produce Haidinger fringe.

How is Haidinger fringe used in scientific research?

Haidinger fringe is used in various scientific research applications, such as studying the properties of light, measuring the accuracy of interferometers, and detecting small changes in the polarization of light. It is also used in optometry to test for certain eye conditions and in geology to identify minerals.

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