Young's double slits - with a twist

[advanced]

Young's double slits - with a twist!

Homework Statement

So this one of the many problems posed in my big book of questions for my current topic (Currently year 12 in England (ie 16-17 year olds) doing AS physics). The question is from a past paper of the British Physics Olympiad (year 2001 Paper 2 Question 1, although no markscheme is available online)

"During a Young's double slit interference experiment, using light of wavelenth 650nm, one slit is covered by a soap film of refractive index 1.35. This causes the zero order fringe to occupy the position previously taken by the 20th fringe. The angle between the directions of the light producing the zero fringe positions is 1.2 degrees. What is the thickness of the soap film and the separation distance of the slits? How would one optically identify these zero order fringes?"

Homework Equations

See https://www.physicsforums.com/library.php?do=view_item&itemid=203 for young's slits equations

and some refractive index equations

The Attempt at a Solution

1) Find d (slit separation), using no film

n*wavelength = d*sin(theta)

d = (20*650*10^(-9))/sin(1.2) ***Is it sin 1.2 or sin 0.6?***

2) Find the soap film thickness

When light passes between two media, velocity and wavelength change but frequency is constant. Also, I need a way of finding a thickness of the film, and I am using light so know it's speed, and can hopefully use speed=distance/time. However I need a time, so thought about expressing the path difference not in distance but as a time, since I can find the time for the original path difference (speed*(d*sin(theta)))=time spent in path difference without the film.

This is where I get stuck... I think I need to link the changing position of the zero order fringe to a new path difference, in terms of a time and then use that to find the thickness of the film.

Also I was wondering whether there could be multiple thicknesses of film that produce fringes in the same position, for instance by adding on a length of film to the existing film corresponding to a distance used up by one full wavelength (or am I missing an obvious point here??)

Thanks anyway - hopefully you can get a bit further than I have!

 

[ideasrule]

Your idea is good: the original time path difference was d*sin(theta)/c, so that has to be the time difference caused by the film. If the film didn't exist, light would take T/c to cross the thickness of the film. However, it does exist, so light would take T/(c/n). The difference between the two is your time difference.

Another way to think about this: in the time light takes to cross the film (T/(c/n)), light travels distance Tn in a vacuum. Tn-T=T(n-1), which has to be equal to the path difference in distance.

 

[vela]

I'd stick with using lengths. The zero-order fringe occurs at the point where there's no difference in optical path length; in other words, the same number of wavelengths are between the source and the screen for the two paths. When the zero-order fringe moves to the n=20 spot, one ray now has a longer path because of the geometry; the other's path is longer because of the soap film.