1. Physics 1B03summer-Lecture 11
Interference of Light
Light is an electromagnetic (EM) wave.
Wave properties:
Diffraction – bends around corners, spreads out from narrow slits
Interference – waves from two or more coherent sources interfere
Physics 1B03summer-Lecture 11

2. Electromagnetic Waves
B (magnetic field) Eo v
Usually we keep track of the electric field E :
Electric field amplitude
Electromagnetic waves are transverse waves
Physics 1B03summer-Lecture 11

3. The Electromagnetic Spectrum
l (m) f (Hz)
3 x
3 x
7 x x1014
4 x 10-7
3 x
3 x
Radio TV
Microwave
Infrared
Visible
Ultraviolet
X rays
g rays
Physics 1B03summer-Lecture 11

4. Physics 1B03summer-Lecture 11
Infrared
780 nm
Red
Yellow 600 nm
Green 550 nm
Blue 450 nm
Violet
380 nm
Ultraviolet l
Visible-Light
Spectrum
Physics 1B03summer-Lecture 11

5. Physics 1B03summer-Lecture 11
Interference: 2 coherent waves, out of phase due to a path difference Dr:
Constructive Interference (maximum intensity)
for f = 0, ±2π, ±4π, ±6π, ………
-> Δr =0, ±l, ±2 l, ±3 l, ………
Destructive Interference (minimum intensity)
for f = ±π, ±3π, ±5π, ………
-> Δr =±λ/2, ±3λ/2, ±5λ/2, ………
Physics 1B03summer-Lecture 11

6. Physics 1B03summer-Lecture 11
Double Slit
(Thomas Young, 1801)
m=2
m=1
m=0 (center)
m=-1
m=-2 θ
incident light
double
slit
separation d
screen
Result: Many bright “fringes” on screen, with dark lines in between.
Physics 1B03summer-Lecture 11

7. Physics 1B03summer-Lecture 11
The slits act as two sources in phase. Due to diffraction, the light spreads out after it passes through each slit. When the two waves arrive at some point P on the screen, they can be in or out of phase, depending on the difference in the length of the paths. The path difference varies from place to place on the screen. P r1
To determine the locations of the bright fringes (interference maxima), we need to find the points for which the path difference Dr is equal to an integer number of wavelengths. For dark fringes (minima), the path difference is integer multiples of half of a wavelength. r2 q d
Δr = r1-r2
Physics 1B03summer-Lecture 11

8. Physics 1B03summer-Lecture 11
For light, the slits will usually be very close together compared to the distance to the screen. So we will place the screen “at infinity” to simplify the calculation.
move P to infinity r1 P
r >> d,
r1 & r2 nearly parallel r2 d q θ
Δr = r1-r2 d θ Δr
Δr = d sin θ
Physics 1B03summer-Lecture 11

9. Physics 1B03summer-Lecture 11
Constructive Interference: (bright)
Δr = mλ or
d sin θ = mλ, m = 0, ±1, ±2, … But, if the slit-screen distance (L) is large, then sinθ~θ
and so sinθ=θ=y/L (in radians): L y θ d
So we have:
Physics 1B03summer-Lecture 11

10. Physics 1B03summer-Lecture 11
Destructive Interference: (no light)
Δr = (m + ½)λ or
d sin θ = (m + ½) λ, m = 0, ±1, ±2, …
So, we have:
Physics 1B03summer-Lecture 11

11. Example y 3 m Where are a) the bright fringes? b) the dark lines?
2 slits, 0.20 mm apart; red light (l = 667 nm) y
3 m
screen
Where are a) the bright fringes?
b) the dark lines?
(give values of y)
Physics 1B03summer-Lecture 11

12. Physics 1B03summer-Lecture 11
Concept Quiz
Which of the following would cause the separation between the fringes to decrease?
Increasing the wavelength
Decreasing the wavelength
Moving the slits closer together
Moving the slits farther apart
None of the above
Physics 1B03summer-Lecture 11

13. Physics 1B03summer-Lecture 11
Concept Quiz
A very thin transparent sheet placed over BOTH slits delays the waves passing through both slits by one-quarter cycle. What happens?
The interference fringes disappear
The fringes all shift upwards
The fringes all shift downwards
There is no change to the pattern.
Physics 1B03summer-Lecture 11

14. Physics 1B03summer-Lecture 11
You want to observe interference with microwaves of frequency 10 Ghz (10 10 Hz). How far apart should you place the slits to see the pattern well?
Does it matter?
Physics 1B03summer-Lecture 11