Young's double slits - with a twist!! 1. The problem statement, all variables and given/known data 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?" 2. Relevant equations See https://www.physicsforums.com/library.php?do=view_item&itemid=203 for young's slits equations and some refractive index equations 3. 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!!!