Presentation on theme: "Electromagnetic Waves G3 Two Source Interference of Waves G4 The Diffraction Grating."— Presentation transcript:
Electromagnetic Waves G3 Two Source Interference of Waves G4 The Diffraction Grating
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.
Single Slit Diffraction (Recap) Central maximum First minimum First subsidiary maximum
Two Source Superposition Experiments Superposition of Sound Waves microphone CRO Signal generator (3kHz) loudspeakers Loud Soft
So in the above example… S 2 Q – S 1 Q = 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 In DB: Topic 4
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.
Path Difference S1S1 S2S2 O Q P Q’ P’ At O : Zero path difference At P and P’ path difference = 1λ At Q and Q’ Path difference = 2λ 2 nd subsidiary maximum 1 st subsidiary maximum Central maximum 1 st subsidiary maximum 2 nd subsidiary maximum
Intensity Distribution Simulation Phet Sim
When slit Width is assumed negligible, fringes are equally bright.
d P S1S1 S2S2 M P’ O D x λ D = distance from slit to screen (few m) d = slit separation (0.5mm) x = distance from central maximum to first subsidiary maximum T θθ
At point P’, the path difference between waves from S 1 and S 2 is one whole wavelength. Thus they arrive in phase, constructively interfere and create a ‘bright fringe’. S 1 T = 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 S 1 TS 2 is a right angle. Using the small angle approximation sin θ = tanθ = θ or... λ = x d D sinθ = λ d tanθ = x D x = Dλ d DB
G4 Diffraction Grating
Effect of increasing the total number of slits (See diagram.) Increasing the total number of slits... - Increases sharpness of fringes - Maxima maintain the same separation - 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.
d B C A θ θ N 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.
Data Booklet It all boils down to three main equations: x = nλx = (n+ ½) λ D d D d Sin θ = nλ d End of presentation
n = 2 Section of grating Monochromatic light n = 1 n = 2 n = 0
Superposition links - PheT Sound (see jar file) - Superposition of two pulsesSuperposition of two pulses - With editable wave equationsWith editable wave equations - Creating a standing waveCreating a standing wave