Unformatted text preview: Interference 5.3 Interference and Diffraction Coherence TwoSlit Interference Multipleslit Interference • Interference effects are a characteristic feature of waves. • Interference of light shows that light has wave properties. • The interference of light waves is shown most clearly when light interacts sm all objects. • Interference effects have many applications, CDs, xray crystallography Coherence For two waves to show interference they must have coherence. Two waves are coherent if one wave has a constant phase rel ati on to the other coherent incoherent Coherence Light from two separat e light bulbs is Incoherent Light from a single sm all source passing through two slits is coherent Laser light is coherent f = 2p Dx l phase shift Interference Interferenc e results from superposition of waves Interference in t wo dimensions black m axim a white m inim a gray zero Sum sum Constructive Interferenc e in phase zero phase shift Coherent waves Interferenc e pattern constructive interference destructive interference Destructive Interferenc e out of phase o phase shift of p, 180 1 Young’s two slit experiment Shows that light shows interferenc e effects. Light has wave properties. Thom as Young dark dark bright bright bright Interference of light waves from the two slits. In the lim it L>>d, the rays are nearly parall el light screen with 2 slits screen light projected on the screen (rotat ed 90o ) Pathlength difference = d sin q Pathlength differences lead to constructive and destructive interference W avelength of light d q Maxima constructive interference Dark destructive interference m is the order of the peak th , i.e m order peak d sin q = m l
d sin q = (m + 1/ 2)l
m=0, + 1, + 2, ............ for sm all angl es q Maxim a Dark y d sin q ; d tan q = d L l y = m L d
1 l y = (m + )L 2d Question Light from a laser is passed through two slits a distance of 0.10 mm apart and is hits a screen 5 m away. The separation between the central m axim um and the first bright interference fringe is 2.6 cm. Find the wavel ength lf the light. Question In a two slit interferenc e experim ent how does the dist ance between the peaks on the screen change if the distance between the slits is increased? a) increases b) decreases c) stays the sam e d) indet erm inate 2 Question In a two slit interferenc e experim ent how does the dist ance between the peaks on the screen change if the wavel ength of the light is increased? a) increases b) decreases c) stays the sam e d) indet erm inate Intensity • When two waves are superimposed the amplitude is equal to the sum of the amplitudes of the two waves. • The intensity of the wave is proportional to the square of the amplitude. Intensity profile The superposition of light from 2 slits gives a sinusoidal intensity profil e (for the case where the slit width is m uch sm aller than d) m=2 m=1 m=0 m=1 m=2 Multiple slits As the num ber of slits having the sam e separation increases the position of the m axim a rem ains the sam e but the width of the peak decreases. 2 slits D 3 slits y 4 slits D/ 2 y 5 slits D/ 3 y D/ 4 y D/5 For multiple slits, destructive interference requires a sm aller phase shift (smaller angle) 1 2 3 Phasor Addition The phasor is a useful tool to describe the superpositi on of sine waves with different phases. A Phasor is specifi ed by an Am plitude and Phase 1/3 cycle j For 3 slits destructive interference is observed when the phase shift is 2p/3 instead of 2p/2 (for 2 slits) Phasors add like vectors 3 Interference two phasors Dj=0 three phasors Dj=0 Intensities due to m ultiple slits m ' l N Destructive interference when m ’ is an integer not a m ultiple of N = no. of slits. Constructive interference when m ’ is an integer m ultiple of N (m =m’/N) d s in q = Dj=p Dj=2p/3 2 slits Dj=2x2p/3 Dj=2p Dj=2p y 4 slits Δ 3 slits D/ 2 Δ/2 5 slits Δ/3 D/3 Δ/4 Δ/5 Diffraction grating • A diffraction grating has a large number of slits N. • As N gets large the width of the peak gets narrow, D/N • The amplitude of the peak increases with N. • The diffraction grating is useful for resolving closely spaced wavelengths of light l Dl l’ Resolution Two different wavel engths of light can be resolved if the peak of one is at the first m inimum of the second. Minim um for mth order l peak = Maxim a for mth order λ’ peak
1 )l = N ' d s inq max = ml ' = m( l + Dl ) d s inq min = ( m + Resolving Power l = mN Dl m order of the diffraction N = no. of slits Spectrometer 4 Question 65 A 400 line/mm diffraction grating is 3.5 cm wide. Two spectral lines whose wavelengths average to 560 nm are just barely resolved in the 4th order spectrum of this grating. W hat is the difference between their wavelengths? 5 ...
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