1. Lecture 16 Diffraction Chp. 37
Topics
Young’s double slit interference experiment
Diffraction and the wave theory
Single slit diffraction
Intensity of single slit diffraction
Circular aperture and double slit diffraction
Diffraction grating
Dispersion and resolving power
Warm-up problem
Demos
Diffraction grating and slits
Inverted mirage
Measuring diameter of a strand Debra’s hair

2. Young’s Double Slit Interference Experimentq
m=0
m=1
m=2 D

3. Maxima Minima m ym +/- 1 2 3 Dl/d 2Dl/d 3Dl/d m ym +/- 1 2 3 Dl/2d1 2 3
Dl/d
2Dl/d
3Dl/d m
ym +/- 1 2 3
Dl/2d
3Dl/2d
5Dl/2d
7Dl/2d

4. What about the intensity of light along the screen?

5. From the table on the previous slide we see that the separation
13E Suppose that Young’s experiment is performed with blue-green light of 500 nm. The slits are 1.2mm apart, and the viewing screen is 5.4 m from the slits. How far apart the bright fringes?
From the table on the previous slide we see that the separation
between bright fringes is

6. Diffraction of a single slit
Find minimum ym
Find maximum
First maximum lies on the axis at q = 0
or ym=0. Other maxima lie in between
the minima. To find them we need to
find the intensity along the screen.

7. Intensity of single slit diffraction

8. Maxima/Minima conditions

9. Exact solution for maxima
To find maxima of a function, take derivative and set equal to 0
Transcendental equation. Solve graphically

11. 8. A 0. 10-mm-wide slit is illuminated by light of wavelength 589nm
8. A 0.10-mm-wide slit is illuminated by light of wavelength 589nm. Consider a point P on a viewing screen on which the diffraction patters of the slit is viewed; the point is at 30o from the central axis of the slit. What is the phase difference between the Huygens wavelets arriving at point P from the top and midpoint of the slit? (Hint: see Eq )
We note that nm = 10-9 m = 10-6 mm. From Eq. 37-4,
This is equivalent to  = 2.8 rad = 160o

12. 6. Sound waves with frequency 3000 Hz and speed 343 m/s diffract through a rectangular opening of a speaker cabinet and into a large auditorium. The opening, which has a horizontal width of 30.0 cm, faces a wall 100 m away. Where along that wall will a listener be at the first diffraction minimum and thus have difficult hearing the sound? (Neglect reflections).
Let the first minimum be a distance y from the central axis which is perpendicular to the speaker. Then q y
Therefore,

13. Diffraction and Interference by a double slit
I = I (double slit interference) x I(diffraction)

14. Sample problem 37-4
How many bright interference fringes fall within the central
peak of the diffraction envelope?
The idea here is to find
the angle where the first
minimum occurs of the diffraction
envelope.
Given
We have m=0 and
m=1,2,3 and 4 on both
sides of central peak.
Ans is 9

15. Diffraction by a circular aperature
The first minimum for the diffraction of light from a circular
aperature is given by:
where d is the diameter of the aperature.
Our ability to resolve two distant point like objects is determined
when the first minimum of one objects diffraction pattern overlaps
the central maximum of another. This is called Raleigh’s criterion.
Example

16. Example 15E. The two headlights of an approaching automobile are 1
Example 15E. The two headlights of an approaching automobile are 1.4 m apart. Assume the pupil diameter is 5.0 mm and the wavelength of light is 550 nm. (a) At what angular distance will the eye resolve them and (b) at what distance?
(a)
(b) s D

17. Diffraction Grating Double slit -- N slits or rulings. w d = w/N
where w is the entire width of the grating

18. Can be used to measure wavelength of light
Measure angles of diffracted lines with a spectroscope using
formula below. Then relate to wavelength
Resolving power of grating.
Measure of the narrowness of lines
Highest R
Highest D
Dispersion of a grating. Measure
of how well lines are separated

19. Show diffraction gratings with increasing N and single slit diffraction with varying slit width a.

20. Babinets Complementarity Principle
In the diffraction region the intensity is the same whether you
have an aperature or opaque disk. You can also replace a slit
with a wire or hair strand. A compact disk is an example of a
diffraction grating in reflection instead of transmission.
Experiment: Measure diameter of a strand of hair from Debra.

21. Mirage
eye
sky
1,09
1.09
1.08
1.08
1.07
1.07
1.06
Hot road causes gradient in the index of refraction that increases
as you increase the distance from the road
In the demo before you the gradient is in the opposite direction

22. Warm-up
HRW6 37.CQ.02. [73994] You are conducting a single-slit diffraction experiment with light of
wavelength l.
(a) What appears, on a distant viewing screen, at a point at which the top and bottom rays through
the slit have a path length difference equal to 5l.?
the m = 5 maximum
the m = 5 minimum
the m = 4 minimum
the m = 4 maximum
(b) What appears, on a distant viewing screen, at a point at which the top and bottom rays through
the slit have a path length difference equal to 4.5l.?
the minimum between the m = 4 and m = 5 minima
the minimum between the m = 4 and m = 5 maxima
the maximum between the m = 4 and m = 5 maxima
the maximum between the m = 4 and m = 5 minima