Double Slit Interference

Double Slit Interference d sin = m  , m=0,1,2,… bright fringe for m=1, need  < d d sin = (m+1/2)  , m=0,1,2,… dark fringe Light gun

16.107 L01 Feb08/02 Problem In a double slit expt the distance between slits is d=5.0 mm and the distance to the screen is D=1.0 m. There are two interference patterns on the screen: one due to light with 1=480 nm and another due light with 2=600 nm. What is the separation between third order(m=3) bright fringes of the two patterns? Note: same d,D but different  => one pattern magnified d/D= 5 x 10-3 => sin ~tan  ~ 

Solution In general d sin  = m and y = D tan since  <<1, ym ~ D m ~ m  D/d for 1=480 nm, D=1.0m, d=5.0 mm y3 = 3(480x 10-9)(1.0)/5.0 x 10-3)= 2.88 x 10-4 m for 2=600 nm, D=1.0m, d=5.0 mm y3 = 3(600x 10-9)(1.0)/5.0 x 10-3)= 3.60 x 10-4 m hence difference is .072 mm

Problem A thin flake of mica (n=1.58) is used to cover one slit of a double slit arrangement. The central point on the screen is now occupied by what had been the seventh bright fringe (m=7) before the mica was used. If = 550 nm, what is the thickness of the mica?

Solution With no mica seventh bright fringe corresponds to path difference = dsin  = 7  with one slit covered, there is an additional path difference due to change of ` = /n in the mica Let unknown thickness of mica be L path difference in mica compared to no mica is [L/ ` - L/ ]  = [nL -L] = [1.58L - L] =.58L to shift seventh bright fringe to center, this must correspond to 7  = .58 L hence L= 7  /(.58) = 7(550)x10-9/.58 = 6.64m

Intensity of Interference Pattern Each slit sends out an electromagnetic wave with an electric field E(r,t)=Emsin(kr-t) where r is the distance from either slit to the point P on the screen the two waves are coherent net field at P is the superposition of two waves which have travelled different distances and hence have a phase difference

Intensity of Interference Pattern net effect at point P is the superposition of two waves E(r,t) = Emsin(kr1-t) + Emsin(kr2- t) =>same wavelength and frequency but travel different distances r1 and r2= r1+ L E(r,t) = Emsin(kr1- t) + Emsin[k(r1+ L)- t] = Em[sin(kr1- t) + sin(kr1- t+ )] where =k L is phase shift due to path difference! sin(A)+sin(B) = 2 cos[(A-B)/2]sin[(A+B)/2] E(r,t) = 2Emcos() sin(kr1- t+ ) ; = /2

E(r,t) = [2Emcos()] sin(kr1-t+ ) =k L =/2=k L/2 L= d sin d k =2/ =/2=d sin / E(r,t) = [2Emcos()] sin(kr1-t+ ) Amplitude= 2Emcos() Intensity (amplitude)2

Intensity of Interference Pattern Amplitude of electric field at P is 2Emcos()= 2Emcos(d sin/) intensity of field  the amplitude squared I = I0 cos2 (d sin/) where I0 =4(Em)2 I = I0 when  =m => d sin=m I = 0 when  = (m+1/2) => d sin=(m+1/2) I = I0 cos2 () with  = d sin/

Intensity of Double Slit E= E1 + E2 I= E2 = E12 + E22 + 2 E1 E2 = I1 + I2 + “interference” <== vanishes if incoherent

MC 41-12 Three coherent equal intensity light rays arrive at P on a screen to produce an interference minimum of zero intensity. If any two of the rays are blocked, the intensity at is I1 . What is the intensity at P if only one is blocked? a) 0 b)I1/2 c) I1 d) 2I1 e) 4I1

Solution y=rsin r  x=rcos y=y1sin(kx-t+1)+ y2sin(kx-t+2)+ y1sin(kx-t+3) Consider adding three vectors of equal length to get zero resultant vector choose 1=0 , 2 = 2/3, 3 = 4/3 add any two => phase difference = 2/3 amplitude = 2y1cos(/2)=2y1cos(/3)=y1 hence intensity is I1 => c)