Diffraction. Factors that affect Diffraction

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1 Diffraction What is one common property the four images share? Diffraction: Factors that affect Diffraction TELJR Publications

2 Young s Experiment AIM: Does light have properties of a particle? Or does it have properties of a wave? This was a big question back in the 19 th and early 20 th centuries and was answered by the results of many famous experiments. The first of these experiments is given below. In 1801 the English scientist Thomas Young ( ) performed an historic experiment that answered the question, is light a wave or is it a particle. He demonstrated this in his famous double slit experiment. Which of the two patterns below would you observe on the screen? Explain. A B The experiment belongs to a general class of "double path" experiments, in which a wave is split into two separate waves that later combine into a single wave. Changes in the path lengths of both waves result in a phase shift, creating an interference pattern. TELJR Publications

3 We will be using the website above to make observations of Young s Experiment. Light Source One Slit Light Source Two Slits Two Light Sources No Slits Two Light Sources Two Slits Based upon the demonstrations, is light exhibit wave properties or particle properties? The bands of light represent Constructive Interference Destructive Interference TELJR Publications

4 What are the conditions for constructive and destructive interference? Definitions: Path Length (l) distance traveled by a wave from a point source to a particular location Path Difference (Δl) difference in path lengths between 2 waves arriving at the same location = l1 - l2 Type of Interference: Type of Interference: Path Length for Source 1 Path Length for Source 2 Path Length for Source 1 Path Length for Source 2 Path Difference Δl Path Difference Δl In general, Constructive Interference occurs when the Path Difference is In general Destructive Interference occurs when the Path Difference is. TELJR Publications

5 Factors that determine the interference pattern (distance between maxima): Factor 1: Factor 2: Factor 3: Formula: TELJR Publications

6 Q.1. Red light ( = 664 nm in vacuum) is used in Young's experiment with the slits separated by a distance d = m. The screen is located at a distance from the slits given by L = 2.75 m. a) Find the displacement of the third order bright fringe from the central bright fringe. b) Find the angle between the central order maxima and the 3 rd order maxima. Q.2. A square is 3.5 m on a side, and point A is the midpoint of one of its sides. On the side opposite this spot, two in-phase loudspeakers are located at adjacent corners. Standing at point A, you hear a loud sound and as you walk along the side of the square toward either empty corner, the loudness diminishes gradually but does not entirely disappear until you reach either empty corner, where you hear no sound at all. Find the wavelength of the sound waves. TELJR Publications

7 Q.3. Blue light of wavelength 440 nm is incident on two slits that are separated by 0.30 mm. Determine the following: a) The angular deflection to the center of the 3 rd order bright band b) The spatial separation from the 0 th order bright band when the light is projected onto a screen that is located 3.0 m from the slits. c) Create a sketch of the above (not to scale) that represents this situation with all known quantities labeled Conclusions: TELJR Publications

8 Diffraction Gratings The diagram to the left shows the interference pattern produced from 2 slits, 4 slits and 16 slits. The y-axis shows the intensity of the light and the x-axis distance from the central maxima (0 th Order maxima). Observations: What is a diffraction grating? Equation for interference patterns produced by a diffraction grating: Q.1. Light from a helium laser has a wavelength of 630 nm and falls upon a diffraction grating. The light then falls on a screen where the first bright spot is separated from the central maximum by 0.51 m. Light of another wavelength of light, falling on the same diffraction grating, produces its first bright spot 0.43 m from the central maximum. Determine the second wavelength. TELJR Publications

9 Single Slit Diffraction AIM: To compare the interference patterns formed from single and double slit diffraction. Diffraction: Q.1. Why do we hear sound even when the source is blocked? Huygen s Principle: Is used to explain diffraction. Each point on the original wavefront in the aperture can be considered to contain many point sources (numbered in the diagram to the right 1 5). This point sources produce smaller wavefronts (called wavelets) that can interfere with each other constructively or destructively depending on path length. Constructive Interference Destructive Interference b Wavelets interfere constructively for bright fringes. Wavelets interfere destructively for dark fringes. TELJR Publications

10 Interference pattern for a Single Slit Interference Pattern for a Double Slit Observations: Observations: Equations: Maxima: Equations: Maxima: Minima: Minima: Q.2. A slit of width a is illuminated by white light. For what value a will the first minimum for red light (λ = 650 nm) appear at an angle of 15 o? Given Need Relation Solution TELJR Publications

11 Q.3. Light passes through a slit and shines on a flat screen that is located L = 0.40 m away. The width of the slit is m. Determine the width of the central bright fringe when the wavelength of the light in a vacuum is λ = 690 nm (red). Conclusions: TELJR Publications

12 Thin Films AIM: How are interference patterns observed in thin films? As you observe the soap bubble you will hopefully see the rainbow of colors. What do you think is causing this effect? If you see a particular color you can assume it is interference. Where is the interference taking place? Look at the diagram below and try to answer this question. When the ray of light strikes the air-soap film boundary some of the light is and some of the light is. The ray that is is then and off of the soap film air bondary. The two rays emerging from the top of the bubble now have an opportunity to interfere. So what determines when it is constructive and destructive for a particular color? Constructive Destructive TELJR Publications

13 Determining when it is constructive or destructive. Ray A When Ray A is incident on the air-soap film boundary, it is partially reflected and partially transmitted. When the wave is reflected at Surface A it is inverted (nair < nsoap ). Ray 1 is a wavelength shifted from the original wave. Air (n= 1.0) Soap Film (n= 1.4) The transmitted wave is not shifted. When the ray reaches Surface B it is NOT inverted (nsoap > nair) It is also not shifted when it emerges out of the film (nsoap > nair). Ray 2 is a wavelength shifted from the original wave. By changing the thickness of the soap film we can create constructive or destructive interference. Q.1. What thickness would the soap film (shown above) need to have in order to have: Constructive Interference Destructive Interference 2t = (m + ½) λ 2t = mλ TELJR Publications

14 Q.1. A soap film (n = 1.33) is 375 nm thick and is surrounded on both sides by air. Sunlight, whose wavelengths (in vacuum) extend from 380 to 750 nm, strikes the film nearly perpendicularly. For which wavelength(s) in this range does constructive interference cause the film to look bright in reflected light? b) What would be the minimum thickness of the film if red light of 700 nm was visible? When Ray A is incident on the air-sio boundary, it is partially reflected and partially transmitted. When the wave is reflected at Surface A it is inverted (nair < nsio ). Ray 1 is a wavelength shifted from the original wave. The transmitted wave is not shifted. When the ray reaches Surface B it is inverted (nsio < nsi) However, it is not shifted when it emerges out of the SiO (nsio > nair). Ray 2 is a wavelength shifted from the original wave. Q.3. What thickness would the soap film (shown above) need to have in order to have: TYPE I TELJR Publications

15 Constructive Interference Destructive Interference 2t = mλ 2t = (m + ½) λ Q.4. Nonreflecting coatings on camera lenses reduce the loss of light at the surfaces of multi-lens systems and prevent internal reflections that might mar the image. Find the minimum thickness of a layer of magnesium fluoride (n = 1.38) on flint glass (n = 1.66) that will cause destructive interference of reflected light of wavelength 550 nm near the middle of the visible spectrum. n = 1 0 n = 1 7 n = 1 2 n = 1 1 n = 1 0 Q.5. Layers of different transparent medium are stacked as shown. When illuminated from above, reflections will occur at the various interfaces. The number of reflected rays experiencing a phase shift is: Conclusions: TELJR Publications

16 Odds and Ends Refraction and dispersion Blue bends best Diffraction and interference Red spreads Soap bubbles Gasoline slicks Camera lenses Glass plates Newton s rings Reflection and interference from two surfaces of thin films TELJR Publications

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