Electromagnetic Waves

G3 Two Source Interference of Waves

G4 The Diffraction Grating

G5 X-Ray Diffraction

Recap: Superposition (interference)

Whenever two waves of the same type meet at the same point, the total amplitude (displacement) at that point equals the sum of the amplitudes (displacements) of the individual waves.

Superposition links

- PheT Sound (see jar file)

- Superposition of two pulses

- With editable wave equations

- Creating a standing wave

Two Source Superposition Experiments

Demo: Superposition of Sound Waves

microphone

CROSignal generator (3kHz) loudspeakers

Path Difference

S1

S2

O

Q

P

Q’

P’

At O : Zero path difference

At P and P’ path difference = 1λ

At Q and Q’ Path difference = 2λ

2nd subsidiary maximum

1st subsidiary maximum

Central maximum

1st subsidiary maximum

2nd subsidiary maximum

So in the above example…

S2Q – S1Q = 2λ

For constructive interference at any point, wavefronts must be ‘in phase’ and their path difference must be a whole number of wavelengths:

path difference = nλ

For destructive interference at any point, wavefronts are ‘π out of phase’ and their path difference is given by:

path difference = (n + ½) λ

Note: The two wave sources must be of similar amplitude for the interference pattern to be observed. If one of the sources is of much greater amplitude, the pattern will not be clearly visible.

Young’s Double Slit Experiment

Setup:

Lamp

Coloured filter

Single slit

Double slit

Screen

Interference pattern

Coherent light

d

P

S1

S2

M

P’

O

D

y

λ

D = distance from slit to screen (few m)d = slit separation (0.5mm) y = distance from central maximum to first subsidiary maximum

T

θθ

At point P’, the path difference between waves from S1 and S2 is one whole wavelength. Thus they arrive in phase, constructively interfere and create a ‘bright fringe’.

S1T = λ

Distance D is large and the wavelength of the light is small so θ must also be very small. We can also approximate that angle S1TS2 is a right angle.

Using the small angle approximation sinθ = tanθ = θ

or...λ = yd D

sinθ = λ d

tanθ = y D

y = Dλd

Effect of increasing the total number of slits

(See diagram.)

Increasing the total number of slits...

- Increases sharpness of fringes

- Gives rise to other faint fringes between the bright fringes

- Increases the intensity of the central maximum

The Diffraction Grating

The diffraction grating is (in theory) a piece of opaque material with many parallel, equidistant and closely spaced transparent slits that transmit light. In practice lines are ruled onto glass with diamond leaving transparent glass in between.

Experiment:Observe a white light source through both coarse (100 lines/mm) and fine (500 lines/mm) gratings. Repeat placing red or green filters in front of the grating.

Observations for white light:

- The diffraction spectra of white light has a central white band (called the zero order image).

- On either side are bright bands of colour. The first red band is the ‘first order spectrum’ for that wavelength. Further out is another identical red band - the ‘second order spectrum’ etc.

- Bands for the visible spectrum are seen, with violet being nearest the centre and red furthest.

- A finer grating forms less orders, further apart than on a coarse grating.

dB

C

Aθ

θN

Section of grating

Light diffracted at θ to the normal

Monochromatic light

Consider a grating with coherent, monochromatic light of wavelength λ incident upon it:

If light from any pair of slits reaching a point in the distance (e.g. focused by the lens in your eye) causes maximum constructive interference, we know that the path difference must be a whole number of wavelengths...

AN = nλ

Hence...

Thus light from all slits constructively interferes with that from every other at values of θ determined by the equation, producing bright regions. This gives rise to the different order spectra for any particular wavelength of light.

d sinθ = nλ n = 0, 1, 2, 3 etc

Q1.

a. Determine the values of θ for the first and second order spectra of light of wavelength 600nm incident upon a fine grating of 600 lines/mm.

b. Which is brightest and why?

c. Explain why a third order spectrum is not possible.

Q2.

Explain (using the formula), why a finer grating produces less orders of spectra than a coarse grating.

n = 2

Section of grating

Monochromatic light

n = 1

n = 1

n = 2

n = 0

Experiment

Use a diffraction grating to determine the wavelength of laser light.

X-Ray Diffraction

Diffracted wavefronts

Incident wavefronts

Plane of atoms

If the diffracted waves are in phase, ABC = nλ

i.e. 2d sinθ = nλ

A

B

C

θ

θ

d

Subtitle

Text

Subtitle

Text

Subtitle

Text

Subtitle

Text

Subtitle

Text