Double Slit Interference

Double Slit Interferenced 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

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 mmy3 = 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 mmy3 = 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 = 6.64m

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

dL= d sin

E(r,t) = [2Emcos()] sin(kr1-t+ )

Amplitude= 2Emcos()

=k L

Intensity (amplitude)2

=/2=k L/2

=/2=d sin /

k =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 + E2

2 + 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

• 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)

rx=rcos

y=rsin

y=y1sin(kx-t+1)+ y2sin(kx-t+2)+ y1sin(kx-t+3)