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Qmechanic ♦
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Is the refractive index neglected in the standard Michelson interferometer equation: lambda=2d$\lambda=2d/NN$

Hoping someone can help with this. It's a simple question, but I can't seem to find the answer anywhere:

I'm looking at the basic Michelson interferometer experiment, where you measure the wavelength of a laser source by changing the relative path lengths, using the moveable mirror.

The equation I keep coming across for this is $\lambda = \frac{2d}{N} $

But, when I try to figure that out for myself, I get the same equation but with the refractive index in there....

My derivation:

OPL = nL

For constructive interference between the two paths on the interferometer, you need

$\Delta OPL = N* \lambda$ (where N is the number of fringes you 'count')

If I change the location of the moveable mirror by length d, then:

$\Delta OPL = n*2d $ (twice d because it traverses the path twice)

So:

$2dn = N\lambda $

and

$\lambda = \frac{2dn}{N} $

However, the 'standard' equation I see on online lab manuals for this is:

$\lambda = \frac{2d}{N} $

Is the n just neglected because it's close to 1, for air? Or is there something deeper here?

Thanks

Is the refractive index neglected in the standard Michelson interferometer equation: lambda=2d/N

Hoping someone can help with this. It's a simple question, but I can't seem to find the answer anywhere:

I'm looking at the basic Michelson interferometer experiment, where you measure the wavelength of a laser source by changing the relative path lengths, using the moveable mirror.

The equation I keep coming across for this is $\lambda = \frac{2d}{N} $

But, when I try to figure that out for myself, I get the same equation but with the refractive index in there....

My derivation:

OPL = nL

For constructive interference between the two paths on the interferometer, you need

$\Delta OPL = N* \lambda$ (where N is the number of fringes you 'count')

If I change the location of the moveable mirror by length d, then:

$\Delta OPL = n*2d $ (twice d because it traverses the path twice)

So:

$2dn = N\lambda $

and

$\lambda = \frac{2dn}{N} $

However, the 'standard' equation I see on online lab manuals for this is:

$\lambda = \frac{2d}{N} $

Is the n just neglected because it's close to 1, for air? Or is there something deeper here?

Thanks

Is the refractive index neglected in the standard Michelson interferometer equation: $\lambda=2d/N$

Hoping someone can help with this. It's a simple question, but I can't seem to find the answer anywhere:

I'm looking at the basic Michelson interferometer experiment, where you measure the wavelength of a laser source by changing the relative path lengths, using the moveable mirror.

The equation I keep coming across for this is $\lambda = \frac{2d}{N} $

But, when I try to figure that out for myself, I get the same equation but with the refractive index in there....

My derivation:

OPL = nL

For constructive interference between the two paths on the interferometer, you need

$\Delta OPL = N* \lambda$ (where N is the number of fringes you 'count')

If I change the location of the moveable mirror by length d, then:

$\Delta OPL = n*2d $ (twice d because it traverses the path twice)

So:

$2dn = N\lambda $

and

$\lambda = \frac{2dn}{N} $

However, the 'standard' equation I see on online lab manuals for this is:

$\lambda = \frac{2d}{N} $

Is the n just neglected because it's close to 1, for air? Or is there something deeper here?

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Hooke
Hooke
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Is the refractive index neglected in the standard Michelson interferometer equation: lambda=2d/N

Hoping someone can help with this. It's a simple question, but I can't seem to find the answer anywhere:

I'm looking at the basic Michelson interferometer experiment, where you measure the wavelength of a laser source by changing the relative path lengths, using the moveable mirror.

The equation I keep coming across for this is $\lambda = \frac{2d}{N} $

But, when I try to figure that out for myself, I get the same equation but with the refractive index in there....

My derivation:

OPL = nL

For constructive interference between the two paths on the interferometer, you need

$\Delta OPL = N* \lambda$ (where N is the number of fringes you 'count')

If I change the location of the moveable mirror by length d, then:

$\Delta OPL = n*2d $ (twice d because it traverses the path twice)

So:

$2dn = N\lambda $

and

$\lambda = \frac{2dn}{N} $

However, the 'standard' equation I see on online lab manuals for this is:

$\lambda = \frac{2d}{N} $

Is the n just neglected because it's close to 1, for air? Or is there something deeper here?

Thanks