Presentation on theme: "Lecture 16 Diffraction Chp. 37"— Presentation transcript:
1Lecture 16 Diffraction Chp. 37 TopicsYoung’s double slit interference experimentDiffraction and the wave theorySingle slit diffractionIntensity of single slit diffractionCircular aperture and double slit diffractionDiffraction gratingDispersion and resolving powerWarm-up problemDemosDiffraction grating and slitsInverted mirageMeasuring diameter of a strand Debra’s hair
3Maxima Minima m ym +/- 1 2 3 Dl/d 2Dl/d 3Dl/d m ym +/- 1 2 3 Dl/2d 123Dl/d2Dl/d3Dl/dmym +/-123Dl/2d3Dl/2d5Dl/2d7Dl/2d
4What about the intensity of light along the screen?
5From 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 separationbetween bright fringes is
6Diffraction of a single slit Find minimumymFind maximumFirst maximum lies on the axis at q = 0or ym=0. Other maxima lie in betweenthe minima. To find them we need tofind the intensity along the screen.
118. 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
126. 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. ThenqyTherefore,
13Diffraction and Interference by a double slit I = I (double slit interference) x I(diffraction)
14Sample problem 37-4How many bright interference fringes fall within the centralpeak of the diffraction envelope?The idea here is to findthe angle where the firstminimum occurs of the diffractionenvelope.GivenWe have m=0 andm=1,2,3 and 4 on bothsides of central peak.Ans is 9
15Diffraction by a circular aperature The first minimum for the diffraction of light from a circularaperature is given by:where d is the diameter of the aperature.Our ability to resolve two distant point like objects is determinedwhen the first minimum of one objects diffraction pattern overlapsthe central maximum of another. This is called Raleigh’s criterion.Example
16Example 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)sD
17Diffraction Grating Double slit -- N slits or rulings. w d = w/N where w is the entire width of the grating
18Can be used to measure wavelength of light Measure angles of diffracted lines with a spectroscope usingformula below. Then relate to wavelengthResolving power of grating.Measure of the narrowness of linesHighest RHighest DDispersion of a grating. Measureof how well lines are separated
19Show diffraction gratings with increasing N and single slit diffraction with varying slit width a.
20Babinets Complementarity Principle In the diffraction region the intensity is the same whether youhave an aperature or opaque disk. You can also replace a slitwith a wire or hair strand. A compact disk is an example of adiffraction grating in reflection instead of transmission.Experiment: Measure diameter of a strand of hair from Debra.
21Mirageeyesky1,091.091.081.081.071.071.06Hot road causes gradient in the index of refraction that increasesas you increase the distance from the roadIn the demo before you the gradient is in the opposite direction
22Warm-upHRW6 37.CQ.02.  You are conducting a single-slit diffraction experiment with light ofwavelength l.(a) What appears, on a distant viewing screen, at a point at which the top and bottom rays throughthe slit have a path length difference equal to 5l.?the m = 5 maximumthe m = 5 minimumthe m = 4 minimumthe m = 4 maximum(b) What appears, on a distant viewing screen, at a point at which the top and bottom rays throughthe slit have a path length difference equal to 4.5l.?the minimum between the m = 4 and m = 5 minimathe minimum between the m = 4 and m = 5 maximathe maximum between the m = 4 and m = 5 maximathe maximum between the m = 4 and m = 5 minima