Change in Optical Path Length

[says]

Homework Statement

A thin film of dielectric material of refractive index 1.455 and thickness 6.50 mm is now placed over one of the slits (say the right-hand one). By how much will this change the optical path between the slit and the centre of the screen?

Homework Equations

Optical Path Difference (OPD) = n(r2-r1)
Where
n = refractive index
L = length between slit and screen
r2 = length light travels from one slit
r1 = length light travels from the other slit

Optical Path Length (OPL) = nL

The Attempt at a Solution

In my lecture notes I've been given the OPD & OPL equation, however I haven't been told what happens to the OPL and thus OPD when light goes through a material of refractive index n.

I've made an attempt, which I've attached. I think I've over thought this question a bit...

 

[ehild]

Homework Statement

A thin film of dielectric material of refractive index 1.455 and thickness 6.50 mm is now placed over one of the slits (say the right-hand one). By how much will this change the optical path between the slit and the centre of the screen?

Homework Equations

Optical Path Difference (OPD) = n(r2-r1)
Where
n = refractive index
L = length between slit and screen
r2 = length light travels from one slit
r1 = length light travels from the other slit

Optical Path Length (OPL) = nL

The Attempt at a Solution

In my lecture notes I've been given the OPD & OPL equation, however I haven't been told what happens to the OPL and thus OPD when light goes through a material of refractive index n.

I've made an attempt, which I've attached. I think I've over thought this question a bit...

Yes, it would have been much simpler to expand the parentheses and simplifying. Your result is almost correct, there is some little calculation error.

 

[says]

Thanks echild!

Can you refer me to the simplification and the calculation error? I've looked at this problem for so long it doesn't make sense anymore!

 

[ehild]

If you type in your derivation, I can show you how to make it simpler.

 

[says]

OPL = n₁*D
OPL₂ = (n₂*t)+(n₁*L₂)

OPD = (N₂*t)+(n₁*L₂)]-(n₁*L₁)
= [(1.455*0.0065m)+(1*5.1235m)] - (1*5.13m)
= 5.132925m - 5.13m
= 2.925 mm

 

[ehild]

Simplify your expression before plugging in the data. Anyway, the result is not 2.925.

OPL = n₁*D
OPL₂ = (n₂*t)+(n₁*L₂)

OPD = (N₂*t)+(n₁*L₂)]-(n₁*L₁)
= [(1.455*0.0065m)+(1*5.1235m)] - (1*5.13m)
= 5.132925m - 5.13m
= 2.925 mm

L₂=L₁-t
OPD = (N₂*t)+n₁*(L₁-t) - (n₁*L₁) = (N₂-N₁)t

 

[says]

Thanks for the simplification, echild!

I plugged in the numbers again and you are right! I appear to have incorrectly calculated [(1.455*0.0065m)+(1*5.1235m)].

 

[says]

This is what I get now.

= [(1.455*0.0065m)+(1*5.1235m)] - (1*5.13m)
= 5.1329575m - 5.13m
= 2.9575 mm

Using simplified equation:

(n₂-n₁)t = (1.455-1)*6.50mm = 2.9575mm

 

[says]

I've now been asked to calculate the new order of interference at the centre of the screen where the zero order originally occurred.

m = (OPD / λ)
m = 2.9575 / 6.33*10^-7
= 4672.2

This doesn't look right to me.

I was thinking that either my OPD is incorrect, or the m value at the original m=0 does equal 4672.2, and this number just represents the number of full cycles (wavelengths).

Therefore, 4672.2 * 633nm = 2.95mm (displacement from original m=0 to new m=0)

I hope this makes sense...

 

[ehild]

They ask how many wavelengths does the optical path difference mean: it is the new order. You calculated it correctly. The thickness of the dielectric is very big, it is not a thin film, it is a very thick slab in optical sense. Maybe, they wanted to say 6.5 micrometer.

 

[says]

It's funny you say that, because the problem statement says the thickness is '6.50 mm'! Might be a typing error.

The big number is confusing me because I thought that maybe I should be using a different equation because the dielectric isn't a 'thin' film. However, I then started to wonder what actually defines a 'thin' film? Isn't it relative to the distance between the slits and the screen. In this questions case, the dielectric is 6.5mm and the distance between the screen and the slits is around 5.13m, therefore would I assume it's a thin film, relative to the distance between the slits and the screen.

I've added a more detailed calculation I did last night.

m = | t/λ_dielectric - t/λ_air |

|absolute value| used to get a positive m value.

n_dielectric*λ_dielectric = n_air*λ_air

t = m * λ_air*n_air/(n_dielectric-n_air)

0.0065 * 0.455 = m * 6.33*10^-7

(0.0065 * 0.455) / 6.33*10^-7 = m

m = 4,672 = new order of interference where the m = 0 order of interference originally was before the dielectric material was placed over one slit.

m * λ = OPD
4672 * 6.33*10^-7 = 2.95 mm

 

[says]

Would the last part of this equation m * λ = OPD = 2.95 mm be the distance where the new m = 0 (order of interference) occurs with the dielectric over the right slit.

I'm assuming the new m = 0 (order of interference) is to the right of the original m = 0 (order of interference) position, because when the light leaves the dielectric it curves away from the normal.

 

[says]

So... I just checked the question and they made an error with the thickness of the dielectric. It's 6.5μm. It's so frustrating that they realized their mistake but then didn't tell anyone about it! I spent a good 5+ hours trying to figure out why my new m = 0 value was over 4,000!

 

[ehild]

Such things happen quite often, micrometer is easy to mix with millimetre. But it is very good that you solved the problem correctly.

 

[says]

Thanks for your help, echild!

Would the last part of this equation m * λ = OPD = 2.95 mm be the distance where the new m = 0 (order of interference) occurs with the dielectric over the right slit.

I'm assuming the new m = 0 (order of interference) is to the right of the original m = 0 (order of interference) position, because when the light leaves the dielectric it curves away from the normal.

 

[ehild]

The new m=0 order occurs when the path difference between the rays emerging from both slits have the same pathlengths.
The beams decline from the normal. You can determine the angle θ when the optical path difference is zero (between the green and blue rays), but it is a bit complicated.
By the way, thin film means that the thickness is comparable to the wavelength. It can be in the micrometer range or less.
So forget that 2.95 mm.