Diffraction at a Single Slit

Transcript

0:00Introduction

Hello everyone! Here is a small puzzle. Light passes through a narrow slit and makes a bright band on a screen. Now make the slit even narrower. The band gets narrower too, right? Well, no, it actually gets wider! Today we find out why, and by exactly how much. Questions on this come up often in JEE, so let's begin.

0:24What is diffraction?

So, what is diffraction? When a wave meets the edges of an obstacle or an opening, it bends round them, and spreads into the region of the geometrical shadow. That is diffraction. Okay? But light usually seems to travel in straight lines, right? So when does this bending become noticeable? When the size of the opening is comparable with the wavelength, yes, exactly! Lambda, the Greek letter on screen, is the wavelength. Clear? In the figure, light comes from the left to a slit whose width is small a, marked at the opening. Every part of the wavefront in the slit acts as a source of secondary wavelets. Their superposition gives the wavy pattern on the right: a bright centre, with weaker bands on both sides. With me?

1:11Dark bands and secondary maxima

Now, where are the dark bands? Theta, the Greek letter in the figure, is the angle from the centre line. The minima are where small a times sine theta equals n times lambda, with n plus or minus one, plus or minus two, and so on; the plus-minus sign gives a dark band on each side. Okay? But n equal to zero is not in the list. Why not? What sits at theta equal to zero? The centre of the pattern, that is correct, and the centre is bright! So the dark bands start at n equal to plus or minus one. Right? Between the dark bands sit the secondary maxima, the weaker bright bands. Here, small a times sine theta is approximately n plus one half, in brackets, times lambda, with n equal to one, two, and so on; the wavy equals sign means approximately. Clear? So the first secondary maximum is at small a times sine theta equals lambda, right? Oh no, careful, that is the first minimum, a dark band! Put n equal to one in our formula, and the first secondary maximum is at about one point five times lambda. Got it?

2:27The central maximum

Now the big bright band in the middle, the central maximum. It runs between the first minima on the two sides. For small angles, sine theta is about theta, so its angular half-width is lambda over small a, and its angular width is two lambda over small a, okay? On a screen at distance capital D, its width is two times lambda times capital D, over small a; that is the arrow labelled two lambda capital D over small a, on the right of the figure. And in the focal plane of a lens of focal length f, it is two lambda f, over small a, right? And this central maximum is twice as wide as every other maximum. Clear?

3:08Brightness, and our narrow-slit puzzle

Now, how bright are the bands? The central and the next secondary maxima have intensities roughly in the ratio one, to four over nine pi squared, to four over twenty five pi squared; pi is the Greek letter. That is one, to zero point zero four five, to zero point zero one six. So the intensity falls very fast. Okay? Now, back to our puzzle. The width is two lambda capital D, over small a, as the badge shows. So if small a gets smaller, does the band get narrower, or wider? Wider, yes, exactly, because small a is in the denominator! So a narrower slit, or a longer wavelength, gives a wider pattern. That is our puzzle solved. Got it? But there is a price. The amplitude at the centre is proportional to small a; the curly sign means proportional to. So the central intensity is proportional to small a squared, and a narrower slit gives a wider band, but a dimmer one. Clear?

4:10Interference and diffraction compared

Now, students often mix up interference and diffraction. In double slit interference, the waves come from two narrow slits; in single slit diffraction, they come from all parts of one slit. Okay? Interference fringes have equal widths and about the same intensity. In diffraction, the intensity falls rapidly from the central maximum. With me? For two slits a distance small d apart, d sine theta equals n lambda gives an interference maximum, a bright fringe. Our single slit formula, small a times sine theta equals n lambda, looks the same, so it is a bright band too, right? No, no, careful, in diffraction it gives a minimum, a dark band! Same shape of formula, opposite meaning. Got it?

4:58Example 1: central maximum width, 500 nm light, 0.1 mm slit

Now example one, an easy one. Light of wavelength five hundred nanometres falls normally on a slit zero point one millimetres wide. Find the width of the central maximum on a screen two metres away. Pause the video and try it first. Done? Okay, let's check. First, everything in metres. Lambda is five times ten to the power minus seven metres, small a is ten to the power minus four metres, and capital D is two metres. Right? Now, the first minimum sits lambda capital D, over small a, from the centre. So is that the width of the central maximum? Not quite, careful here, that is only the half-width! There is a dark edge on each side, so the width is two lambda capital D, over small a, which is two times five times ten to the power minus seven, times two, over ten to the power minus four. With me? On top, we get two times ten to the power minus six. Dividing by ten to the power minus four gives two times ten to the power minus two metres, which is twenty millimetres. See our puzzle again: a slit only zero point one millimetres wide makes a band twenty millimetres wide! Clear?

6:18Example 2: fringes inside central maximum, missing 5th order

Now a harder one. Example two: in a double slit arrangement, each slit has width zero point one millimetres, and the slits are zero point five millimetres apart. How many bright interference fringes lie inside the central diffraction maximum? Pause the video and try it. Here is a hint: each slit is also a single slit. Done? Alright, let's go. Each slit gives a single slit pattern, and the fringes sit inside its central maximum. It ends at the first diffraction minimum, where small a times sine theta equals lambda. Okay? The interference maxima are at d sine theta equals n lambda. At that edge, sine theta is lambda over small a, so n equals small d over small a; the double arrow means so. That is zero point five over zero point one, which is five. With me? So the orders run from minus five to plus five, which makes eleven bright fringes, right? No, no, careful, that is the trap here! The fifth order falls exactly on the first diffraction minimum, so that bright fringe is missing. Got it? So we see orders zero, plus or minus one, and so on up to plus or minus four. That is two times small d over small a, minus one, which gives nine fringes. Clear?

7:43Quick revision: minima, central maximum width, intensity

So, quick revision. Minima are at small a sine theta equals n lambda, with n plus or minus one, two, and so on, and secondary maxima near n plus one half, times lambda. The central maximum is two lambda over small a in angle, and two lambda capital D over small a on a screen, twice as wide as the others. And a narrower slit gives a wider pattern, but the central intensity, proportional to small a squared, drops. Now go and try the practice problems on this page. All the best!