Optics question help. Find Optical path difference

Sorry if this doesn't quite meet 'undergraduate' standard, but I didn't do A-level physics so have no idea if it was covered there or not.

The context of the question is a Michelson interferometer, the distance between the fixed mirror and splitter is equal to the distance between the adjustable mirror and splitter.

2L_2 - 2L_1 = m\lambda

where L_n is the splitter to mirror length.

given that the (L_2) = (L_1) ass told in the question, and that

\lambda

is a wavelength in nm also given in the question, then m must equal zero. Thus allowing me to set subsquent equations to zero and solve for r_1, which is then used to work out the OPD.

There is an object of thickness 6mm and refractive index of 1.55 between the fixed mirror and splitter.

The question asks what is the optical path difference of the resulting fixed arm?

What I currently have so far:

I worked out the optical path length (n*r),

OPL = n_{1}\times{r_{1}}

where:

n = refractive index = 1.55
r = distance travelled through the medium = 6mm

I also worked out the wavelength in the medium.

I have no idea how to obtain a value for the OPD though. // update: think I've maybe got an answer but any comments are still appreciated as I have no way to check my answer really.

Using the equation for OPD:

OPD = (n_{2}\times{r_{2}}) - (n_{1}\times{r_{1}})

where [latex]n_1 \times r_{2}[\latex] = refractive index of a vacuum (=1) x distance though the other medium, which I think should be the total distance from the splitter to the fixed mirror.

I haven't been given any values other than the refractive index of the medium and the thickness of the medium, so I have no value for r_1 and not sure how find it. All I know about the length is that the distance from the two mirrors to split are the same

So to update this I think I have worked it out, but I'm not sure. I rearranged various equations around to get the value of r_1 which I can use to work out the OPD. When I use that r_1 value to solve for n_2, n_2 is equal to the value for n_2 given in the question. I'm unsure though if this is because I worked it correctly, or I worked it incorrectly but consistently, which all my values fit into equations perfectly but produce madly incorrect answers. One hand wavy area is that I think OPD equation above might also equal m, so setting m to 0 might not make sense, or maybe I'm just too tired right now

(edited 8 years ago)

Reply 1

Et Tu, Brute?

OP

21

looking like I'm on my own for this one, though I think I might have got it now anyway.

Can anyone tell me though what a reasonable answer for a change in optical path would be in a Michelson interferometer? I have a negative value in the x10^-3 region.

Reply 2

Et Tu, Brute?

OP

21

@Smack
@a10

Sorry to try rope you guys into some physics, but you're the only people I know on this site who might have come across this, I know you guys are doing/did mechanical engineering, maybe optics was covered?

Sorry, can't help with this. Might have a second look tomorrow to see if I can figure out anything based on the units involved but I have absolutely no knowledge of optics.

Reply 4

Joinedup

20

do you mean the compensator?

that's there to compensate for the thickness of the beam splitter - light on the fixed mirror path is reflected off the surface of the beam splitter whereas light on the translatable mirror path is transmitted through the thickness of the beam splitter (which is typically made of glass)

Reply 5

Et Tu, Brute?

OP

21

Original post by Joinedup

do you mean the compensator?

that's there to compensate for the thickness of the beam splitter - light on the fixed mirror path is reflected off the surface of the beam splitter whereas light on the translatable mirror path is transmitted through the thickness of the beam splitter (which is typically made of glass)

The question just mentioned an object placed between the splitter and the fixed mirror, which would be the compensator yes.

Reply 6

Joinedup

20

Original post by Et Tu, Brute?

The question just mentioned an object placed between the splitter and the fixed mirror, which would be the compensator yes.

Well I'd say that the optical path difference between the interferometer without the object inserted and with the object inserted would have to be 2n₁r₁ because the light beam passes through the object twice...

Reply 7

Et Tu, Brute?

OP

21

Original post by Joinedup

Well I'd say that the optical path difference between the interferometer without the object inserted and with the object inserted would have to be 2n₁r₁ because the light beam passes through the object twice...

good point. Why would i need to times the refractive index by 2 also though? Couldn't I just times the distance it travels by 2, ie

2r_1

?

Reply 8

Et Tu, Brute?

OP

21

Another update:Pretty sure I am actually wrong, so frustrating.

OPD =\frac{r_2}{\lambda_2} - \frac{r_{1}}{\lambda_1} =(\frac{n_2r_2}{\lambda_0}) - (\frac{n_1r_1}{\lambda_0}) = m'

firstly, not sure what the difference in m and m' are.Secondly, I have lambda0,1 and 3 here, suggesting 3 waves of light. Or maybe landba0 is meant to be lamba1. Anyway, if I set m' to 0, then it means that r2 = r1 which surely cannot be the case.So only slightly better off that I was this time last night. Any help is appreciated.

Reply 9

Joinedup

20

Original post by Et Tu, Brute?

good point. Why would i need to times the refractive index by 2 also though? Couldn't I just times the distance it travels by 2, ie

2r_1

?

yeah you're multiplying the thickness by 2, not the refractive index. it's the same material but the beam goes through two times the thickness of it on a 2 way journey.

the change in optical path length for a beam is n₁r₁ each time it goes through... so if it goes through twice it's two lots of that...

2n₁r₁

Original post by Et Tu, Brute?

Another update:Pretty sure I am actually wrong, so frustrating.

OPD =\frac{r_2}{\lambda_2} - \frac{r_{1}}{\lambda_1} =(\frac{n_2r_2}{\lambda_0}) - (\frac{n_1r_1}{\lambda_0}) = m'

firstly, not sure what the difference in m and m' are.Secondly, I have lambda0,1 and 3 here, suggesting 3 waves of light. Or maybe landba0 is meant to be lamba1. Anyway, if I set m' to 0, then it means that r2 = r1 which surely cannot be the case.So only slightly better off that I was this time last night. Any help is appreciated.

not sure what you're trying to answer here tbh.

Reply 10

Et Tu, Brute?

OP

21

Original post by Joinedup

not sure what you're trying to answer here tbh.

I've made a bit of a hash of this question I think.

I'm trying to work out the change in optical path length of the fixed arm. So I am guessing that is the difference in path length with the material and the path length without it (ie. just vacuum).

Without knowing the vale for r1 (distance the light travels through the medium, the medium being the vacuum between fixed mirror (M1) and the beam splitter (BS), I don't know how to compare them.

I know the vacuum (n1) has a refractive index of 1, r2 is the thickness of the material between BS and M1 and n2, refractive index of that material is also given as well as wavelength of the source beam.

From this I can workout optical path length of the material and the wavelength of light in it.

I can't workout the change in optical path length of the fixed arm however.

I feel that it involves m, which I think is the order/number of fringes. I found an equation which I can calculate the difference in m, but got something insane like 9000, so pretty sure that is wrong.