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UMass Amherst PHYSICS 132 - physics 132 lab report 1

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Diffraction of light using variables , d, y, m, and LVictoria NewburyPhysics 132 - Section Number: MLHasbrouck 212Lab Section instructor: Zachary MeadowsDate due: 2/7/14Abstract:In this experiment we used different types of lens in order to diffract the light from a laser. The laser was fixed and connected to a magnetic holder. The magnetic holder will hold the different types of lenses in place. The lenses we will be using are a straight edge/opaque disk, focal lens, a disk with several narrow slits, a double slit disk, and a diffraction gradient. We used the light produced on the wall in order to measure variables , d, y, m, and L. We altered variables during the process in order to see how this impacted the other variables. Questions and answers:1. Define wavelength and frequency in your own words and state the relation between them (equation statement).a. Wavelength is the distance between the peaks of waves that come one after the other. Frequency is the number of cycles per second and is measured in hertz. When wavelength and frequency are multiplied, they equal the speed of light in air (3.0 x 108). Light can travel through materials other than air, in this case thewavelength and speed can change but the frequency always stays the same. The type material will affect how the wavelength or speed changes.2. Describe the pattern observed from diffraction around a straight edge, for the straight edge/opaque disks. Why do you think this pattern is diffraction and not interference?a. This pattern is diffraction and not interference because of the resulting light on thewall. The light is shown in the “shadow region” on the wall despite that the straight edge is blocking part of the laser. This observation shows that the light is diffracting around the edge and that its waves are bending around the end and spreading. This is a characteristic of diffraction. During interference, light waves are added together. Interference produces a pattern of dark and bright areas. Waves that are constructive, or in phase, allow you to see a light. Waves that are destructive, or out of phase, won’t produce light.3. When you illuminate the opaque holes in the straight edge/opaque disk using the laser through the lens, was there a light or dark spot at the center of the screen?a. There was a light spot at the center of the screen when the laser illuminated the opaque holes in the straight edge/opaque disk using the laser through the lens.4. Is there a specific reason for using lasers in this experiment, and not ordinary white light, if yes, what is it (there may be more than one reason)?a. The reason for using lasers in this experiment rather than ordinary white light is because a beam of light behaves like a wave. White light, on the other hand, travels like a stream of particles. The beam of light is collimated (diverges very little) so it easier to measure light points on the wall. These properties allow the measurement of certain variables as wavelength, frequency, angles of light, “a” (width of the slit), w (length of the central peak), L (distance from light source to the wall), d (distance between slits), y (distance from center to middle of mth band), and r (distance from center of slits to a point on the wall).5. For single slit diffraction how would the width of the central maxima or the central bright band change a) if the wavelength increased, or b) if the slit width “a” is increased?a. If the wavelength increased, the central bright band would…b. If the slit width “a” increased, the central bright band would…6. Pretend you redid the diffraction grating section and arrived at this data: L = 1m, d = 3.33 x 10-4 cm, y (distance from center to first bright spot) = 22.1 cm, =656 nm (red light)… Now pretend you leave the equipment setup the same but pass blue light ( = 396 nm) through the diffraction grating. What y distance will the blue light be diffracted to? Is this closer to or farther away fromthe center?a. m *  = (d*y) / L  1 * 6.56 * 10-5 cm = (3.33 * 10-4 cm * y) / 1 m  y = .197 cmb. Replace  = 656 nm with  = 396 nmi. 1 * 3.96 * 10-5 cm = (3.33 * 10-4 * y) / 1 m  y = .119 cmc. Based on the above calculations, the y distance the blue light will be diffracted to be .119 cm. This is closer to the center.7. A light spectrum is formed on the screen using a diffraction grating. The entire apparatus made up of laser, grating and the screen is now immersed in a liquid with refractive index 1.33. Do the bright spots on the screen get closer together, farther apart, remain the same or disappear entirely? Explain.a. The bright spots on the screen will get closer together because the wavelength of the laser decreases when it goes through the liquid. The wavelength impacts the distance between the bright spots because there will be a smaller angle of diffraction due to the reduced wavelength.8. Light with wavelength of  = 456 nm is incident on a diffraction grating. The screen is placed at a distance L = 2 m from the diffraction grating. The grating has N = 200/mm parallel slits (used to calculate d). Find the distance y between the 2nd bright spots or the second order maxima on the either side of the central bright spot.a. (dy)/L = mb.  = 4.56* 10-7 m, m = 2, L =2 m, d = 5 * 10-9 m; need to find y.c. y = (mL)/d = (2 * 4.56 * 10-7 * 2)/ (5 * 10-9) = 364.8d. y = 364.8Conclusion:Using the formula (dy)/L = m, it is possible to calculate the outcomes of light patterns on the wall. This also makes it possible to predict how the pattern would change given different variables. Our results were not completely accurate, this was in part due to the difficulty in measuring the distances between lights as they are very small and it can be difficult to tell where is the center of the light and where the end of it


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UMass Amherst PHYSICS 132 - physics 132 lab report 1

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