Understanding the wave nature of light through diffractionAtreyi SahaPhysics 132 - Section LLMondays 8:00 AMHasbrouck 212February 8, 2016AbstractIn this experiment, we investigated the diffraction of light. Diffraction is a type of interference. First, we observed how light emitted from a helium-neon laser is diffracted by a straight edge and disk. Then, we examined how the laser light was diffracted by single narrow slit and a double slit. We then used a diffraction grating device to measure the wavelength of the helium-neon laser and compared that value to our wavelength measurements obtained from the single and double slit experiments,Questions & Answers1. Define wavelength and frequency in your own words and state the relation between them,(equation statement). (0.5 point)Wavelength refers to the amount of distance between the two peaks of a wave. Frequencyrefers to how fast a given wave vibrates or goes up and down. Frequency and wavelengthshare an inverse relationship where a wave that has a high frequency also has a short wavelength and vice versa.2. Describe the pattern observed from diffraction around a straight edge. Based on your observation with the straight edge experiment, do you conclude light is a wave or a particle? Why?The light that appeared on the screen was stretched out horizontally. From this, I conclude that light is a wave because waves bend around sharp edges. When waves comein contact with a sharp edge or corner, they will bend and spread out as circular waves. The horizontally stretching light I observed was a result of this bending and spreading phenomenon of waves described above thereby evidencing that light is a wave.3. When you illuminate the opaque disk 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? Why?(1.0 point)There was a black spot shaped like a ring at the center of the disc. In the middle of this black ring, there were tiny bright spots. The black spot at the dead center of disc is most likely due to light waves cancelling out, an example of destructive interference. The tiny bright spots are due to minor instances of wave superposition, an example of constructiveinterference.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) (1.5 points)Yes, there are multiple reasons why we used a helium-neon laser instead of ordinary white light. Firstly, the light emitted from a laser is monochromatic unlike ordinary white light which is a combination of many different wavelengths. The advantage of using a monochromatic light source is that it is one wavelength or color. Additionally, the light emitted from a laser is coherent which means that the wavelengths of the laser light are inphase both in time and in space. 5. For single slit diffraction how would the width of the central maxima or the central brightband changeFormula used to solve the problem: y/L = m(wavelength)/aa) If the wavelength is increased (1.0 point) If the wavelength is increased, then the central maximum width would also increase. This is because the central maximum is directly proportional to wavelength.b) If the slit width “a” is decreased (1.0 point)If the slit width is decreased, then the central maximum width would increase because slit width and the central maximum are inversely related. 6. Pretend you redid the diffraction grating section and arrived at this data: L = 1m, d = 3.33x 10 -4 cm, y (distance from center to first bright spot) = 22.1cm, λ=656nm (red light).Now pretend you leave the equipment setup the same but pass blue light (λ=452nm) through the diffraction grating. What y distance will the blue light be diffracted to? Is thiscloser to or farther away from the center? (2.5 point)y/L =[ m (wavelength)] /dRed light0.221/1 = m(6.56 x 10-7) / (3.33 x 10-6)m = 1.12y = 0.221 m Blue lighty/1 = (1.12)(4.52 x 10-9) / (3.33 x 10-6)y = 1.53 x 10-3 mEverything aside from the wavelength remains unchanged. In this case, wavelength is reduced and since wavelength is directly proportional to the distance from the central maximum (y), the y distance becomes smaller with the blue light and is therefore close 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 index1.5. Do the bright spots on the screen get closer together, farther apart, remain the same or disappear entirely? Explain (1.5 points)Because light goes from air (n =1) to the liquid (b = 1.5), the light will slow down. This causes the wavelength to be reduced and a shorter wavelength means that the diffracted rays will interfere constructively and reach full wavelength path differences given a smaller angle of diffraction. From this, we can conclude that the bright spots will move closer together and there will be more of them located at the outside edge of the 8. Light with wavelength of λ = 500 nm is incident on a diffraction grating. The screen is placed at a distance L = 2m from the diffraction grating. The grating has N = 500/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. (2.0 point)y/L =[ m (wavelength)] /dd = 1/500000 y/2 = [2 (5x10-7)] (500000)y = 1 x 2y = 2.00 mConclusionWe were able to use our observations about how light emitted from a helium-neon laser is diffracted by a straight edge and disk to confirm the wave nature of light. We then used three tools to measure the wavelength of the laser light. The actual wavelength of the helium-neon laser light is 633 nm. First, by using single-slit diffraction, we determined the wavelength of the helium-neon laser light to be 692 nm. This is very similar to the actual wavelength value with only a 9.3% error. Next, by using double-slit diffraction, we determined the wavelength of the helium-neon laser light to be 481 nm. This is very different from the actual wavelength value with a 24% error. Finally, by using single-slit diffraction, we determined the wavelength of the helium-neon laser light to be 550 nm. This is also similar to the actual wavelength value with only a 13.3% error.Possible sources of error include errors in measurement. These can arise from limits in
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