Shift of Fringes and Thin Films

Transcript

0:00Introduction

Hello everyone! Here is a puzzle. Picture a double slit, with bright and dark fringes on a screen. Cover one slit with a clear sheet of mica, much thinner than a hair. Light still passes through, yet the whole pattern slides sideways! Which way, and by how many fringes? And why does a thin oil layer on water shine in colours? Both come up often in JEE papers. So, let's begin.

0:28A thin sheet over one slit

So, take a thin transparent sheet, of thickness t and refractive index mu, the Greek letter on screen, and place it over one slit. Okay? Inside the sheet, light covers a distance t. Optical path is the index times the distance, so the sheet gives mu times t, where air gives just t, right? So how much extra optical path does the sheet add? Yes, exactly, mu times t, minus t! That is, mu minus one, in brackets, times t. Clear? Now the big question. The central bright fringe sits where the two optical paths are equal, and the covered beam has extra path. So does the central fringe move towards the covered slit, or away? Towards the covered slit, very good! There, the path in air from that slit is shorter, which makes up for the sheet. The shift, capital delta y, the Greek delta with a y, is mu minus one, times t, times capital D, over small d. Capital D is the screen distance, and small d is the slit gap. With me?

1:37Count the shift in fringes

Now count it in fringes. For small angles, the fringe width beta, the Greek letter on screen, is lambda times capital D, over small d, with lambda the wavelength. Got it? Multiply and divide capital delta y by lambda, and it becomes mu minus one, times t, over lambda, times beta. The double arrow on screen means therefore. So the pattern shifts by mu minus one, times t, over lambda, fringes. Okay? Now, here is a trap. The sheet makes one path longer. So do the fringes also get wider, or narrower? No, no, the width stays the same! Beta depends only on lambda, capital D and small d, and the sheet changes none of them. Right? And in a liquid of index mu l, that is mu with a small l below, each arrow means replace with. Replace mu minus one with mu over mu l, minus one, and replace lambda with lambda over mu l. Clear?

2:38Thin films: two reflected rays

Now, thin films, in the figure. The grey strip is a film of index mu and thickness t. Ray one reflects from the top; ray two refracts into the film, travels inside at angle r, reflects from the bottom, and comes out. The faint arrow is light going through. Both rays come from one incoming ray, so they are coherent. Okay? Ray two travels down and back up inside the film. Compare it with ray one, and the optical path difference works out to two mu t cos r, where cos r is the cosine of the angle r. The note beside the figure says so too. Got it? And the note adds lambda over two for each reflection at a denser medium. Such a reflection gives a phase change of pi, the Greek letter on screen, which is the same as half a wavelength of extra path. With me? But at a rarer medium, there is no phase change at all. So at each reflection, ask: is the other side denser, or rarer? Right?

3:37Bright or dark in reflected light

Now take a soap film in air. At the top, light in air reflects off soap; at the bottom, light in soap reflects off air. Which reflection gives the phase change? Only the top one, correct, since soap is denser than air! So ray one already carries half a wavelength extra, and the film must supply the other half, plus any whole number of wavelengths. It is bright when two mu t cos r equals n plus one half, times lambda, for n equal to zero, one, two and so on. Okay? Now tell me this: if two mu t cos r equals n times lambda, a whole number of wavelengths, is the reflection bright or dark? It is dark, very good! The extra half wavelength makes the two rays cancel. That is the opposite of the double slit, a classic trap. Clear? When both reflections have the phase change, or neither has it, like a coating with index between air and glass on a glass lens, the conditions are swapped. Got it? And the transmitted pattern is complementary to the reflected one: where the reflection is bright, the transmitted light is dim. Right?

4:46Example 1: fringe shift by a 6 micrometre mica sheet

Now, back to our mica sheet. Example one: a mica sheet of thickness six micrometres and index one point five is placed over one slit. By how many fringes does the pattern shift for light of wavelength six hundred nanometres? Pause the video and try it first. Done? Okay, let's check. The mu m means micrometre, ten to the power minus six metre, and careful, this mu is a unit, not the index! And n m means nanometre. So t is six times ten to the power minus six metre, and lambda is six times ten to the power minus seven metre. With me? Mu minus one is zero point five. So the number of fringes is zero point five times six times ten to the power minus six, over six times ten to the power minus seven. Okay? On top, zero point five times six gives three. And three times ten to the power minus six, over six times ten to the power minus seven, is exactly five. Five whole fringes towards the covered slit, and our puzzle is solved! Clear?

5:53Example 2: strongest and missing colour from an oil film

Now a harder one. Example two: white light falls normally on an oil film of index one point four and thickness three hundred nanometres, floating on water of index one point three three. Which visible wavelength is strongest in the reflected light, and which is missing? Pause the video and try it first. Done? Alright, let's go. Normal incidence means r is zero, so cos r is one. Then two mu t is two times one point four times three hundred, which is eight hundred and forty nanometres. Okay? Now, watch out for this. At the bottom, light in the oil reflects off water, and water is denser than air, so there is a phase change there too, right? No, no, compare oil with water, not air with water! Water's one point three three is less than oil's one point four, so water is rarer here, and the bottom has no phase change. Only the top, from air to oil, has it. Got it? Bright needs two mu t to equal n plus one half, times lambda, so lambda is eight hundred and forty, over n plus one half. For n equal to zero, one and two, that gives one thousand six hundred and eighty, then five hundred and sixty, then three hundred and thirty six nanometres. With me? Dark needs two mu t to equal n times lambda, so lambda equals eight hundred and forty over n, which gives eight hundred and forty, then four hundred and twenty nanometres. Okay? Visible light is about four hundred to seven hundred nanometres. So green, at five hundred and sixty, is the strongest, and violet, at four hundred and twenty, is missing. That is why thin oil films on water look coloured. Clear?

7:34Quick revision: fringe shift, 2 mu t cos r, phase change

So, quick revision. Sheet over one slit: a shift of mu minus one, times t, over lambda, fringes, towards that slit, with beta unchanged. Thin film: path difference two mu t cos r, plus half a wavelength per reflection at a denser medium. With one phase change: bright at n plus one half, times lambda, and dark at n times lambda. Now try the practice problems on this page. All the best!