1. Physics 7E
Prof. D. Casper

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Reminder: Programmable/graphing calculators are NOT allowed
Chapter 34 HW due tomorrow morning
Chapter 35 HW due Monday morning
Review session finalized: Saturday, 2 – 4 pm, ICS 174
Monday
Tuesday
Wednesday
Thursday
Friday
Saturday
Today
Ch. 34 HW Due
Ch. 35 Disc.
Wrap-up
Review
ICS 174
2-4 pm
Ch. 35 HW Due
Final (8am)

3. Two sources (separated by 1 𝜆)

4. Examining the interference pattern
We know that constructive interference (antinodes) will occur when 𝑟 1 − 𝑟 2 =𝑚 𝜆 and destructive interference (nodes) will occur when 𝑟 1 − 𝑟 2 = 𝑚+ 1 2 𝜆 (where 𝑚 is an integer) We can graph the places where these conditions occur

5. Q35.5
Two radio antennas radiating in phase are located at points A and B, which are 6 wavelengths apart. A radio receiver is moved along a line from point B to point C. A 6l B C
At what distances from point B will the receiver detect an intensity maximum?
4.5l 8l
C. 9l
D. both A. and B.
E. all of A., B., and C.
Answer: D

6. A35.5
Two radio antennas radiating in phase are located at points A and B, which are 6 wavelengths apart. A radio receiver is moved along a line from point B to point C. A 6l B C
At what distances from point B will the receiver detect an intensity maximum?
Constructive interference will occur when
Δ𝑟= 𝑟 𝐴 − 𝑟 𝐵 =𝑚𝜆
The only chance is if we have something similar
to a 3:4:5 right triangle
For a 6:8:10 triangle, Δ𝑟=10𝜆−8𝜆=2𝜆
For a 4.5:6:7.5 triangle, Δ𝑟=7.5𝜆−4.5𝜆=3𝜆
4.5l 8l
C. 9l
D. both A. and B.
E. all of A., B., and C.

7. Young’s two-slit interference experiment
In 1800, Thomas Young performed a famous experiment that demonstrated interference of light (and first measured the wavelength of red light)

8. Analyzing Young’s experiment
To find the locations of constructive (bright)
and destructive (dark) interference on the screen we need to calculate the difference in distance
of a point 𝑃 from the two slits
Make a simplifying approximation:
assume the distances to the screen ( 𝑟 1 and 𝑟 2 ) are much greater than the distance 𝑑 between slits.
Then the rays can be treated as parallel to each other,
and the difference in distance at some angle 𝜃 is:
Δ𝑟= 𝑟 1 − 𝑟 2 =𝑑 sin 𝜃

9. Location of bright bands
With the simplification, we have constructive interference when Δ𝑟=𝑑 sin 𝜃 =𝑚 𝜆 If the screen is a distance 𝑅 from the slits, the lateral position 𝑦 𝑚 of the 𝑚 𝑡ℎ bright band is given by 𝑦 𝑚 𝑅 = tan 𝜃 𝑚 For small angles (only!), tan 𝜃 𝑚 ≈ sin 𝜃 𝑚 ≈ 𝜃 𝑚 , so: 𝑦 𝑚 =𝑅 𝑚𝜆 𝑑

10. The Small-Angle Approximation
If the screen distance 𝑅 is large compared to the lateral distance 𝑦, then
sin 𝜃 ≈ tan 𝜃 ≈ 𝑦 𝑅 R

11. Spacing of bands?
If we increase the spacing between the slits (sources), will the spacing between the bright bands on the screen:
Increase
Decrease
Stay the same

12. Effect of source spacing
𝑑=2𝜆
𝑑=𝜆

13. Intensity of the Interference Pattern
We found the location of the maxima and minima
Now we’ll find the intensity of the interference pattern at any point
Three step process:
Find the amplitude of the electric field at point P
𝐸 𝑃 = 𝐸 1 + 𝐸 2 from superposition principle
Depends on phase difference 𝜙 between 𝐸 1 and 𝐸 2 at point P
Find the intensity of the wave
Intensity proportional to amplitude squared: 𝐼= 1 2 𝜖 0 𝑐 𝐸 𝑃 2
Relate the phase difference 𝜙 to the position of P
𝜙=𝑘( 𝑟 1 − 𝑟 2 ) if waves emitted in phase

14. Amplitudes and Intensities
Q: Why can’t we just add the intensities? A: Because we have coherent sources that interfere Adding intensities would be the correct thing to do for incoherent sources and would lead to NO interference pattern
Adding intensities
(incoherent sources)
Adding amplitudes
(coherent sources)

15. Applying the Superposition Principle
Assume we have waves from two sources that differ only by a phase at point P: 𝐸 1 (𝑡)=𝐸 cos(𝜔𝑡+𝜙) 𝐸 2 𝑡 =𝐸 cos 𝜔𝑡 (We assume the maximum amplitudes 𝐸 are equal, ignoring the dependence on distance) The superposition principle says the electric field at P is: 𝐸 𝑃 𝑡 = 𝐸 1 𝑡 + 𝐸 2 𝑡

16. Solving for 𝐸 𝑃
Graphically, we can use the Law of Cosines to find the magnitude of the resultant electric field: 𝐸 𝑃 2 = 𝐸 2 + 𝐸 2 −2 𝐸 2 cos 𝜋−𝜙 Trig identity: cos (𝜋−𝜙) =− cos 𝜙 𝐸 𝑃 2 =2 𝐸 2 (1+ cos 𝜙 ) Trig identity: cos 𝜙 2 = 1+ cos 𝜙 2 𝐸 𝑃 =2𝐸 cos 𝜙 2

17. From Amplitude to Intensity
From Chapter 32: 𝐼= 1 2 𝜖 0 𝑐 𝐸 𝑃 2 Intensity is proportional to the maximum amplitude ( 𝐸 𝑃 ) squared Using 𝐸 𝑃 =2𝐸 cos 𝜙 2 : 𝐼= 1 2 𝜖 0 𝑐 4 𝐸 2 cos 2 𝜙 2 ≡ 𝐼 0 cos 2 𝜙 2

18. Phase Difference and Position
Each wavelength difference in distance corresponds to 2𝜋 difference in phase: 𝜙 2𝜋 = 𝑟 2 − 𝑟 1 𝜆 Therefore 𝜙= 2𝜋 𝜆 𝑟 2 − 𝑟 1 ≡𝑘( 𝑟 2 − 𝑟 1 ) If 𝑟 1 , 𝑟 2 ≫𝑑, then as we saw earlier, 𝑟 2 − 𝑟 1 ≈𝑑 sin 𝜃 𝜙≈ 2𝜋𝑑 𝜆 sin 𝜃

19. Intensity Function
Combining everything: 𝐼= 𝐼 0 cos 2 𝜋𝑑 𝜆 sin 𝜃 If the transverse distance 𝑦≪𝑅, then sin 𝜃 ≈𝑦/𝑅 and 𝐼≈ 𝐼 0 cos 2 𝜋𝑑𝑦 𝜆𝑅