The Diffraction Grating - SFU.ca · 2010. 1. 15. · The Diffraction Grating If one extends the...

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The Diffraction Grating

If one extends the double slit to largenumber of slits very closely spaced, onegets what is called a diffraction grating.d sinθ. Maxima are still at

d sinθm = mλ,m = 0,1,2,3, . . .

The difference is that the fringes arethinner and brighter.

Neil Alberding (SFU Physics) Physics 121: Optics, Electricity & Magnetism Spring 2010 1 / 17

Lines of high intensity occur onlywhere the wavefronts from all the slitsinterfere constructively. Therefore themaxima are very intense and verynarrow.

The angle from the middle of thegrating to the maxima is given by

d sinθm = mλ,m = 0,1,2,3, . . .

The distance from the centralmaximum to the next maximum isgiven by

ym = L tanθm

Lines of high intensity occur onlywhere the wavefronts from all the slitsinterfere constructively. Therefore themaxima are very intense and verynarrow.The angle from the middle of thegrating to the maxima is given by

d sinθm = mλ,m = 0,1,2,3, . . .

ym = L tanθm

Lines of high intensity occur onlywhere the wavefronts from all the slitsinterfere constructively. Therefore themaxima are very intense and verynarrow.The angle from the middle of thegrating to the maxima is given by

d sinθm = mλ,m = 0,1,2,3, . . .

ym = L tanθm

The angles to the maxima are notsmall. Therefore, the small angleapproximation cannot be used. Thedistance on the screen to the brightlines is given by

ym = L tan�sin−1

�mλd

��

The distances to the maxima providea good way of determiningwavelengths of light.Diffraction gratings are essentialcomponents of optical spectrometers.

�mλd

��

The distances to the maxima providea good way of determiningwavelengths of light.

Diffraction gratings are essentialcomponents of optical spectrometers.

�mλd

��

The distances to the maxima providea good way of determiningwavelengths of light.Diffraction gratings are essentialcomponents of optical spectrometers.

Why are the fringes thinner and brighter?The brightness comes from the fact that constructive interferenceis now happening from more sources:

Imax = N2I1

One way to think about the narrowness is conservation of energy.If the bright fringes will be so much brighter (gain a square in theformula above) then they must be narrower too.

Why are the fringes thinner and brighter?The brightness comes from the fact that constructive interferenceis now happening from more sources:

Imax = N2I1

One way to think about the narrowness is conservation of energy.If the bright fringes will be so much brighter (gain a square in theformula above) then they must be narrower too.

Why are the fringes thinner and brighter?Another way to think of it is to look at the geometry again. Imagine weintroduce a small phase difference between adjacent slits, say 1.1λinstead of 1λ

With two adjacent slits, this does not affect the intensity very muchNow imagine the interference of 2 slits which are separated by 5d(non-adjacent slits). 5 × 1.1λ = 5.5λ. Now these two slits aredestructively interfering!Total width of the central max is (for large number of slits)

2θmin =2λNd

Measuring θmin can be another way to determine an unknown λ

Reflection Gratings

Many common gratings are actuallyreflection gratings rather thantransmission gratings.A mirror with thousands of narrowparallel grooves makes a gratingwhich reflects light instead oftransmitting it, but the math is thesame.A CD is an excellent example.

Compact Discs

A CD is a reflective surface coveredin small bumps/pits. Each second ofmusic requires ∼ 106 bumps.

From above there are “pits” in analuminum layer, but CD is read frombelow, so “bumps” are seen.A CD player shines a laser on thebumps in a spiral track and detectsthe reflected laser beam.The intensity of the reflected beamchanges as the bumps and landspass by. Height of a bump is aboutλ/4, making a path difference of λ/2and a phase change of π.

Compact Discs

A CD is a reflective surface coveredin small bumps/pits. Each second ofmusic requires ∼ 106 bumps.From above there are “pits” in analuminum layer, but CD is read frombelow, so “bumps” are seen.

A CD player shines a laser on thebumps in a spiral track and detectsthe reflected laser beam.The intensity of the reflected beamchanges as the bumps and landspass by. Height of a bump is aboutλ/4, making a path difference of λ/2and a phase change of π.

Compact Discs

A CD is a reflective surface coveredin small bumps/pits. Each second ofmusic requires ∼ 106 bumps.From above there are “pits” in analuminum layer, but CD is read frombelow, so “bumps” are seen.A CD player shines a laser on thebumps in a spiral track and detectsthe reflected laser beam.

The intensity of the reflected beamchanges as the bumps and landspass by. Height of a bump is aboutλ/4, making a path difference of λ/2and a phase change of π.

Compact Discs

A CD is a reflective surface coveredin small bumps/pits. Each second ofmusic requires ∼ 106 bumps.From above there are “pits” in analuminum layer, but CD is read frombelow, so “bumps” are seen.A CD player shines a laser on thebumps in a spiral track and detectsthe reflected laser beam.The intensity of the reflected beamchanges as the bumps and landspass by. Height of a bump is aboutλ/4, making a path difference of λ/2and a phase change of π.

Compact Discs Tracking System

A diffraction grating is used.The central maximum readsthe bumps while the1st-order maxima are usedas tracking beams (theyshould not fluctuate).

Single Slit Diffraction

It is rather strange to talkabout thousands of slits beforetalking about 1. However,thousands are actually a littleeasier.

A single slit diffraction patterninvolves a wide centralmaximum flanked by weakersecondary maxima and darkfringes.It would appear that we haveonly one light source in thiscase, so how do weunderstand the interference?We have to go back toHuygen’s principle.

It is rather strange to talkabout thousands of slits beforetalking about 1. However,thousands are actually a littleeasier.A single slit diffraction patterninvolves a wide centralmaximum flanked by weakersecondary maxima and darkfringes.

It would appear that we haveonly one light source in thiscase, so how do weunderstand the interference?We have to go back toHuygen’s principle.

It is rather strange to talkabout thousands of slits beforetalking about 1. However,thousands are actually a littleeasier.A single slit diffraction patterninvolves a wide centralmaximum flanked by weakersecondary maxima and darkfringes.It would appear that we haveonly one light source in thiscase, so how do weunderstand the interference?

We have to go back toHuygen’s principle.

It is rather strange to talkabout thousands of slits beforetalking about 1. However,thousands are actually a littleeasier.A single slit diffraction patterninvolves a wide centralmaximum flanked by weakersecondary maxima and darkfringes.It would appear that we haveonly one light source in thiscase, so how do weunderstand the interference?We have to go back toHuygen’s principle.

A wave front passes through a narrowslit (width a). Note that narrow isimportant.

Each point on the wave-front emits aspherical waveOne slit becomes the source of manyinterfering wavelets.A single slit creates a diffractionpattern on the screen.

A wave front passes through a narrowslit (width a). Note that narrow isimportant.Each point on the wave-front emits aspherical wave

One slit becomes the source of manyinterfering wavelets.A single slit creates a diffractionpattern on the screen.

A wave front passes through a narrowslit (width a). Note that narrow isimportant.Each point on the wave-front emits aspherical waveOne slit becomes the source of manyinterfering wavelets.

A single slit creates a diffractionpattern on the screen.

A wave front passes through a narrowslit (width a). Note that narrow isimportant.Each point on the wave-front emits aspherical waveOne slit becomes the source of manyinterfering wavelets.A single slit creates a diffractionpattern on the screen.

Why the Wide Central Maximum?

Wavelets from any part of the slithave to travel approximately the samedistance to reach the center of thescreen.

A set of in-phase wavelets thereforeproduce constructive interference atthe center of the screen.

Why the Wide Central Maximum?

Wavelets from any part of the slithave to travel approximately the samedistance to reach the center of thescreen.A set of in-phase wavelets thereforeproduce constructive interference atthe center of the screen.

Why the Dark Bands?

Consider the path-lengths well awayfrom the centre axis

For any wavelet it is possible to find apartner which is a/2 away.If the path difference betweenpartners happens to be λ/2 then thispair will create total destructiveintereference. A dark band will becreated.For any given wavelength there willbe an angle for which this condition istrue! There will always be darkbands, as long as a is greater than λand the slit is narrow.

Why the Dark Bands?

Consider the path-lengths well awayfrom the centre axisFor any wavelet it is possible to find apartner which is a/2 away.

If the path difference betweenpartners happens to be λ/2 then thispair will create total destructiveintereference. A dark band will becreated.For any given wavelength there willbe an angle for which this condition istrue! There will always be darkbands, as long as a is greater than λand the slit is narrow.

Why the Dark Bands?

Consider the path-lengths well awayfrom the centre axisFor any wavelet it is possible to find apartner which is a/2 away.If the path difference betweenpartners happens to be λ/2 then thispair will create total destructiveintereference. A dark band will becreated.

For any given wavelength there willbe an angle for which this condition istrue! There will always be darkbands, as long as a is greater than λand the slit is narrow.

Why the Dark Bands?

Consider the path-lengths well awayfrom the centre axisFor any wavelet it is possible to find apartner which is a/2 away.If the path difference betweenpartners happens to be λ/2 then thispair will create total destructiveintereference. A dark band will becreated.For any given wavelength there willbe an angle for which this condition istrue! There will always be darkbands, as long as a is greater than λand the slit is narrow.

The Mathematics of the Dark Bands

The path difference between 1 and 2 is

∆r12 =a2

sinθ1 =λ2

What about the other angles for destructiveinterference? The general formula becomes

a sinθp = pλ,p = 1,2,3, . . .

The small angle approximation means wecan write

θp = pλa,p = 1,2,3, . . .

But if a is small then θp is large and thesmall angle approximation is not valid.

∆r12 =a2

sinθ1 =λ2

a sinθp = pλ,p = 1,2,3, . . .

θp = pλa,p = 1,2,3, . . .

∆r12 =a2

sinθ1 =λ2

a sinθp = pλ,p = 1,2,3, . . .

θp = pλa,p = 1,2,3, . . .

The Width of the Bands

It can be useful to express the fringeposition in distance rather than angle.

The position on the screen is given byyp = L tanθp. This leads to

yp =pλL

a,p = 1,2,3, . . .

The width of the central maximum is giveby twice the distance to the first darkfringe

w =2λL

aIt is important to note that: 1) the widthgrows if the screen is farther away 2) Athinner slit makes a wider centralmaximum.

It can be useful to express the fringeposition in distance rather than angle.The position on the screen is given byyp = L tanθp. This leads to

yp =pλL

a,p = 1,2,3, . . .

w =2λL

It can be useful to express the fringeposition in distance rather than angle.The position on the screen is given byyp = L tanθp. This leads to

yp =pλL

a,p = 1,2,3, . . .

w =2λL

Circular Aperture Diffraction

θ1 =1.22λ

And the width of the central maximum is

w = 2y1 = 2L tanθ1 ≈2.44λL

Circular Aperture Diffraction

θ1 =1.22λ

DAnd the width of the central maximum is

w = 2y1 = 2L tanθ1 ≈2.44λL

The Wave and Ray Models of Light

The factor that determines how mucha wave spreads out is λ/a

With water or sound we seediffraction in our everyday livesbecause the wavelength is roughlythe same as the macroscopicopenings and structures we seearound us.We will only notice the spreading oflight with apertures of roughly thesame scale as the wavelength of light.

The factor that determines how mucha wave spreads out is λ/aWith water or sound we seediffraction in our everyday livesbecause the wavelength is roughlythe same as the macroscopicopenings and structures we seearound us.

We will only notice the spreading oflight with apertures of roughly thesame scale as the wavelength of light.

The factor that determines how mucha wave spreads out is λ/aWith water or sound we seediffraction in our everyday livesbecause the wavelength is roughlythe same as the macroscopicopenings and structures we seearound us.We will only notice the spreading oflight with apertures of roughly thesame scale as the wavelength of light.

Sometimes we treat light like a stream of particles, sometimes like awave and sometimes like a ray.

Does light travel in a straight line ornot? The answer depends on the circumstances.

Choosing a Model of LightWhen light passes through openings < 1mm in size, diffractioneffects are usually important. Treat light as a wave.When light passes through openings > 1mm in size, treat it as aray.

Sometimes we treat light like a stream of particles, sometimes like awave and sometimes like a ray. Does light travel in a straight line ornot?

The answer depends on the circumstances.

Sometimes we treat light like a stream of particles, sometimes like awave and sometimes like a ray. Does light travel in a straight line ornot? The answer depends on the circumstances.

The Diffraction Grating - SFU.ca · 2010. 1. 15. · The Diffraction Grating If one extends the...

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