Diffraction of Light: Single Slit, Resolution and the Grating – Interference and Diffraction of Light (Senior 5 Physics) | Ganzaa

Interference and Diffraction of Light

Lessons

Senior 5 · Physics

Interference and Diffraction of Light

4 lessons

Lesson 4 of 4

Diffraction of Light: Single Slit, Resolution and the Grating

25 min

Light that bends round corners

Hold two fingers almost touching in front of your eye and look through the narrow gap at a distant lamp: faint dark lines appear, running parallel to your fingers. Look at a street light through the fine weave of an umbrella and you see a grid of bright spots. Tilt a CD under a lamp and it throws out every colour of the rainbow. All three are diffraction: the spreading of a wave when it passes through a narrow opening or past an edge. Diffraction sets a limit to how sharp any image can be, in your eye, a camera or a telescope, and it is used in the spectrometers that identify the chemical elements in stars.

Diffraction and Huygens' principle

When a plane wave meets an opening, every point of the opening acts, by Huygens' principle, as a source of wavelets. If the opening is much wider than the wavelength, the wavelets rebuild a plane wave and the beam goes straight on with a sharp-edged shadow. If the opening is comparable with the wavelength, the wavelets spread out in all directions and the wave bends into the shadow region. Light's wavelength is under a micrometre, so its diffraction is noticeable only with very narrow openings or over long distances, which is why everyday shadows look sharp.

Diffraction by a single slit

Light of wavelength $\lambda$ passes through a slit of width $a$ onto a distant screen. Consider the direction $\theta$ in which the path difference between the wavelets from the two edges of the slit is exactly $\lambda$ : $a\sin\theta = \lambda$ . Divide the slit into an upper and a lower half. Each wavelet from the upper half has a partner in the lower half, a distance $\frac{a}{2}$ below it, whose path is longer by $\frac{a}{2}\sin\theta = \frac{\lambda}{2}$ . Every pair cancels, so this direction is dark. Dividing the slit into 4, 6, … parts gives further dark directions:

a\sin\theta = n\lambda, \qquad n = \pm 1, \pm 2, \pm 3, \ldots \quad \text{(minima)}

Note the difference from the double slit: here the formula gives dark directions, and $n = 0$ is excluded because the centre is bright. For a screen at distance $D$ , the minima are at $y = \frac{n\lambda D}{a}$ , so the central maximum has width $\frac{2\lambda D}{a}$ , twice that of the side maxima. The full intensity pattern is $I = I_0\left(\frac{\sin\beta}{\beta}\right)^2$ with $\beta = \frac{\pi a\sin\theta}{\lambda}$ :

Narrowing the slit makes the pattern wider: the more you try to confine light, the more it spreads.

Definition

Diffraction is the spreading of a wave as it passes through an opening or round an obstacle; it is significant when the opening is comparable with the wavelength. For a single slit of width $a$ , the dark fringes are in directions given by $a\sin\theta = n\lambda$ ( $n = \pm 1, \pm 2, \ldots$ ). Example: for $a = 0.10$ mm and $\lambda = 633$ nm, the first minimum is at $\sin\theta = 6.3 \times 10^{-3}$ , an angle of 0.36°.

Worked example 1: a laser through a slit

Red light of wavelength 633 nm from a laser passes through a slit 0.10 mm wide onto a wall 2.0 m away. Find the width of the central maximum. What happens if the slit is narrowed to 0.050 mm?

Solution

Width $= \frac{2\lambda D}{a} = \frac{2(633 \times 10^{-9})(2.0)}{0.10 \times 10^{-3}} = 2.5 \times 10^{-2}$ m $= 25$ mm.

Halving $a$ doubles the width, to about 51 mm, and the pattern becomes dimmer, since less light gets through and it is spread more widely.

Diffraction and resolution

A circular opening of diameter $b$ , such as the pupil of your eye or the lens of a telescope, diffracts light into a bright central disc surrounded by faint rings. The first dark ring is at

\sin\theta = \frac{1.22\lambda}{b}

So the image of a point of light is never a point but a small blurred disc. Two point sources close together give two overlapping discs. By the Rayleigh criterion, they can just be distinguished (resolved) when the centre of one disc falls on the first dark ring of the other, that is, when their angular separation is

\theta_{\min} \approx \frac{1.22\lambda}{b} \quad \text{(in radians)}

A larger aperture or a shorter wavelength gives finer detail. That is why astronomers build telescopes with huge mirrors.

Worked example 2: how far can you see two headlights?

At night the pupil of the eye is about 3.0 mm across. Taking $\lambda = 550$ nm, find (a) the smallest angle the eye can resolve, (b) the greatest distance at which the two headlights of a car, 1.5 m apart, are seen as two separate lights.

Solution

(a) $\theta_{\min} = \frac{1.22(550 \times 10^{-9})}{3.0 \times 10^{-3}} = 2.2 \times 10^{-4}$ rad.

(b) For small angles, separation $= \theta \times$ distance, so distance $= \frac{1.5}{2.24 \times 10^{-4}} = 6.7 \times 10^{3}$ m, about 7 km. Beyond that the two headlights merge into one. (In practice the eye's limit is somewhat worse, because of imperfections in its lens.)

The diffraction grating

A diffraction grating is a plate ruled with a very large number of equally spaced parallel slits (or grooves): typically 300 to 1200 per millimetre. If there are $N$ lines per metre, the spacing is $d = \frac{1}{N}$ . Light from neighbouring slits reinforces in directions where the path difference between them is a whole number of wavelengths:

d\sin\theta = n\lambda, \qquad n = 0, \pm 1, \pm 2, \ldots \quad \text{(maxima)}

This is the same condition as for a double slit, but with thousands of slits contributing, the light from them cancels almost completely except in these exact directions. The maxima are therefore very sharp and bright, and their angles can be measured precisely. Since $\sin\theta \le 1$ , the highest order possible is the largest whole number not greater than $\frac{d}{\lambda}$ .

White light is spread into a spectrum in each order, with violet nearest the centre, because $\theta$ increases with $\lambda$ . A grating spectrometer measures the angles of the lines in a spectrum, and so their wavelengths, which identifies the elements present: this is how the composition of the Sun and the stars is known. The closely spaced tracks of a CD (1.6 μm apart) act as a reflection grating.

Definition

A diffraction grating is a large number of equally spaced parallel slits, spacing $d$ ; bright maxima occur where $d\sin\theta = n\lambda$ , and $n$ is the order. Example: for 600 lines per mm, $d = \frac{1}{6.00 \times 10^{5}} = 1.67 \times 10^{-6}$ m.

Worked example 3: sodium light through a grating

Sodium light (589 nm) falls normally on a grating with 600 lines per millimetre. Find (a) the angles of the first- and second-order maxima, (b) the highest order that can be seen.

Solution

$d = \frac{1}{600 \times 10^{3}} = 1.667 \times 10^{-6}$ m.

(a) $\sin\theta_1 = \frac{589 \times 10^{-9}}{1.667 \times 10^{-6}} = 0.353$ , $\theta_1 = 20.7^\circ$ . $\sin\theta_2 = 0.707$ , $\theta_2 = 45.0^\circ$ .

(b) $\frac{d}{\lambda} = \frac{1.667 \times 10^{-6}}{589 \times 10^{-9}} = 2.83$ , so the highest order is 2. A third order would need $\sin\theta = 1.06$ , which is impossible.

Warning

Same-looking formulas, different meanings. Double slit and grating: $d\sin\theta = n\lambda$ gives bright maxima, with $d$ the slit spacing. Single slit: $a\sin\theta = n\lambda$ gives dark minima, with $a$ the slit width, and $n \ne 0$ . For a grating, remember to convert "lines per mm" to the spacing $d$ in metres.

Tip

Narrow spreads wide. Every diffraction formula has the opening in the denominator: a smaller slit, pupil or line spacing always gives a wider pattern.

Diffraction grating, Khan Academy. Watch how many equally spaced slits produce sharp maxima at angles given by d sin θ = nλ.

✦

Practice

Explain why diffraction of sound round a doorway is easy to notice, but diffraction of light is not.
Light of wavelength 500 nm passes through a slit 0.20 mm wide. Find the angle of the first minimum and the width of the central maximum on a screen 1.5 m away.
Explain, using the idea of pairs of wavelets, why the direction $a\sin\theta = \lambda$ is dark.
A telescope has an objective 0.20 m in diameter. Find the smallest angle it can resolve at 550 nm.
A grating has 400 lines per mm. Find the angle of the second-order maximum for light of 650 nm, and the highest order visible.
The first-order maximum for a certain light through a grating of 500 lines per mm is at 17.5°. Find the wavelength.
White light passes through a grating. Explain why each order is a spectrum, and which colour is deviated most.
A student argues that a grating with more lines per mm is always better for measuring wavelengths. Evaluate this, considering both the spread of the spectrum and the number of orders.

Show answersHide answers

Sound wavelengths (about 0.3 m to 3 m) are comparable with a doorway, so the spreading is large. Light wavelengths (under 1 μm) are about a million times smaller than everyday openings, so the spreading is tiny.
$\sin\theta = \frac{500 \times 10^{-9}}{0.20 \times 10^{-3}} = 2.5 \times 10^{-3}$ , $\theta = 0.14^\circ$ ; width $= \frac{2(500 \times 10^{-9})(1.5)}{0.20 \times 10^{-3}} = 7.5$ mm.
Split the slit into two halves. Each wavelet in the top half pairs with one $\frac{a}{2}$ lower, whose path is longer by $\frac{a}{2}\sin\theta = \frac{\lambda}{2}$ ; each pair cancels, so the total is zero.
$\theta = \frac{1.22(550 \times 10^{-9})}{0.20} = 3.4 \times 10^{-6}$ rad.
$d = 2.5 \times 10^{-6}$ m; $\sin\theta_2 = \frac{2(650 \times 10^{-9})}{2.5 \times 10^{-6}} = 0.52$ , $\theta_2 = 31^\circ$ ; $\frac{d}{\lambda} = 3.85$ , highest order 3.
$d = 2.0 \times 10^{-6}$ m; $\lambda = d\sin\theta = (2.0 \times 10^{-6})\sin 17.5^\circ = 6.0 \times 10^{-7}$ m $= 600$ nm.
Each wavelength has its own angle, $\sin\theta = \frac{n\lambda}{d}$ , so the colours separate in every order except zero. Red, with the longest wavelength, is deviated most; violet least.
Partly true. More lines per mm (smaller $d$ ) spreads the spectrum over larger angles, so angles, and hence wavelengths, can be measured more precisely. But fewer orders are then available ( $n_{\max} \le \frac{d}{\lambda}$ ), and long wavelengths may not appear at all beyond first order. The best grating gives a large spread while still showing the orders needed.

Key points

Diffraction: spreading of waves through openings, significant when the opening is comparable with $\lambda$ .
Single slit: minima at $a\sin\theta = n\lambda$ ( $n \ne 0$ ); central maximum $\frac{2\lambda D}{a}$ wide, twice the others.
Circular aperture: $\theta_{\min} = \frac{1.22\lambda}{b}$ (Rayleigh criterion) limits resolution.
Grating: sharp maxima at $d\sin\theta = n\lambda$ , $d = \frac{1}{N}$ , $n_{\max} \le \frac{d}{\lambda}$ ; white light gives spectra.

1Light that bends round corners

2Diffraction and Huygens' principle

3Diffraction by a single slit

4Diffraction and resolution

5The diffraction grating

Light that bends round corners

Diffraction and Huygens' principle

Diffraction by a single slit

Diffraction and resolution

The diffraction grating