INTERFERENCE OF em WAVES in 2D (ch. 27 in text)
Visible light:   400 nm  700 nm

To see interference need:
o Closely spaced sources with constant phase relation → coherent
 Independent sources (i.e. lightbulbs) are incoherent → no
interference
YOUNG’S DOUBLE-SLIT EXPERIMENT (1801)
 Observation of interference in light → demonstrated wave character
Light through a slit: Diffraction

Plane wave incident on slit

Slit acts like source of circular waves
Light through two slits: Interference

Get constructive interference along lines where paths from each slit to point
on line differ by nλ
o along this line, crests from both slits overlap
r1
r2
Constructive
for r2 - r1= n
SCREEN ILLUMINATED BY DOUBLE SLIT:
 see pattern of bright and dark fringes

bright where waves from slits arrive in phase
(Δpath = nλ)

dark where waves from slits arrive out of
phase (Δpath = (n+1/2)λ)

spacing of bright spots depends on
o wavelength
o slit separation
o distance from slits to screen
PATH DIFFERENCE AND INTERFERENCE (27.3 in text)
P
Illuminated
slits
r1
y
r2
λ
d
s2
screen

s1
Path difference = r2 – r1
L

For constructive interference: r2  r1  m

For destructive interference: r2  r1  m  1/ 2

where
m  0,1, 2, 3,
where
m  0,1, 2, 3,
If L  d (normal case), then paths ~ parallel
o Both paths make same angle  with line perpendicular to plane of slits
MAGNIFY REGION NEAR SLITS: (will find  by similar triangles)
r1
r2
d


  r2  r1  d sin 
Now write conditions for constructive/destructive interference in terms of angle:

Constructive interference for d sin  bright  m

1

d
sin


m



bright
Destructive interference for
2


m  0, 1, 2, 3,
m  0, 1, 2, 3,
P
Illuminated
slits
From the angle, we can
calculate POSITIONS of
bright fringes on the screen
λ
r1
r2

s1
screen
Path difference = r2 – r1
d s2
L

y
tan 
L

y
sin


But for small , have tan  sin  so that
L (OK for large L, small y)

So from d sin  bright  m ,

m
y

L
Result for location of fringes: bright
d
get
d
y bright
L
 m
m  0, 1, 2, 3,
o Says that for screen far from slits, bright fringes are equally spaced
o Gives us a way to measure wavelength of light
y
EXAMPLE: fringes from a red laser shining through a pair of slits
λ = 638 nm

Red laser:

Experiment: L (slits to screen) = 5.0 m,

m
y

L
From bright
d , find separation of adjacent (Δm = 1) bright fringes is:
d (slit separation) = 0.2 mm
9
 1   5.0 m  638 10 m
ybright  L
 0.016 m

-3
d
0.2

10
m


Alternatively:
 If wavelength is known, can use this to measure small distance between
slits.
P
Illuminated
slits
λ
INTENSITIES IN A TWO-SLIT
INTERFERENCE PATTERN:
r1
d s2
Path difference = r2 – r1
L
For arbitrary point P at angle :





y
screen
r2

s1
Path difference for waves reaching P from the 2 slits is r2  r1  d sin 
2
2





r

r

d sin 
2
1
Resulting PHASE DIFFERENCE is


2   d sin  
I

I
cos


max
Average Intensity of light at P is average



dsin
Maxima at d sin   m
Bright/dark fringes of
equal width
2λ
2-slit pattern
λ
λ

0
d
λ
intensity
2λ
DIFFRACTION GRATING:
For many slits, separated by d, maxima still at d sin   m BUT much narrower
dsin
Diffraction grating
λ
2λ
λ

0
d
λ
2λ
intensity
Can make diffraction grating by cutting hundreds of closely-spaced grooves on
piece of glass or plastic
APPLICATIONS OF DIFFRACTION GRATING: SPECTROMETER
Device to separate
and measure
wavelengths of light
from excited atoms,
stars etc.

Telescope
(for viewing and
measuring  )
 red
source
collimator
 blue
grating
m
sin


From
d , see that
o For red → longer wavelength → get larger 
o For blue → shorter wavelength → get smaller 

Measuring wavelengths gives info about:
o energy levels in atoms and molecules
o composition of gases, unknown samples, even stellar atmospheres
o motion of stars and galaxies from Doppler shift of wavelength