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Sight and wave phenomena Topic A.4 Diffraction For this section, we will be dealing with Diffraction at a single slit When a He-Ne beam passes through a single narrow slit, this was observed on a distant screen What happened? Recall that when waves meet an aperture or barrier, they diffract. The image shows the diffraction of the light at the slit. Take a closer look at the image, what can you observe? Did you observe 2 things? The centre spot is the brightest, and the spots get dimmer as they go to the sides The centre spot is about twice the length of the spots adjacent to it. Sketch the variation with angle of diffraction of the relative intensity of light diffracted at a single slit. Single Aperture Diffraction Pattern With a narrower aperture, the overall intensity of the light is reduced and the pattern on the screen is wider. This is how the image and the graph coincide Diffraction Patterns When plane wavefronts pass through a small aperture they spread out This is an example of the phenomenon called diffraction Light waves are no exception to this However, when we look at the diffraction pattern produced by light we observe a fringe pattern, that is, on the screen there is a bright central maximum with "secondary" maxima either side of it. There are also regions where there is no illumination and these minima separate the maxima. This intensity pattern arises from the fact that each point on the slit acts, in accordance with Huygen's principle, as a source of secondary wavefronts. It is the interference between these secondary wavefronts that produces the typical diffraction pattern. Derive the formula θ = λ / b for the position of the first minimum of the diffraction pattern produced at a single slit. Consider the diagram below In particular we consider the light from one edge of the slit to the point P where this point is just one wavelength further from the lower edge of the slit than it is from the upper edge. The secondary wavefront from the upper edge will travel a distance λ/2 further than a secondary wavefront from a point at the centre of the slit. Hence when these wavefronts arrive at P they will be out of phase and will interfere destructively. The wavefronts from the next point below the upper edge will similarly interfere destructively with the wavefront from the next point below the centre of the slit. In this way we can pair the sources across the whole width of the slit. If the screen is a long way from the slit then the angles θ1 and θ2 become nearly equal. From the previous figure we see therefore that there will be a minimum at P if sin θ1 = λ / b where b is the width of the slit. However, both angles are very small, and θ1≈ θ2 ≈ θ, so we can write that θ =λ/b With this, we have linked the size of the slit to the angle and wavelength of the wave. What is the relationship between the size of the image formed on the screen and the wavelength? tan θ ≈ y/D tan θ ≈ sin θ ≈ θ ≈ y/D b There is a mistake in the diagram, can you spot it? If the distance of the screen to the slit is D, the actual width of the maximum along the screen can be calculated: y θ D tan θ = y/D For very small values of θ, we approximate tan θ ≈ θ Therefore θ ≈ y/D , since θ = λ / b We have y = λ D / b Solve problems involving single-slit diffraction. Example Light from a laser is used to form a single slit diffraction pattern. The width of the slit is 0.10 mm and the screen is placed 3.0 m from the slit. The width of the central maximum is measured as 2.6 cm. What is the wavelength of the laser light? Answer Since the screen is a long way from the slit we can use the small angle approximation. The central max = 2.6 => y = 1.3 x 10-2m Given D = 3.0m, b = 0.10 x 10-3m Using y = λ D / b λ = (1.3 x 10-2) x(0.10 x 10-3) / 3.0 λ = 430 x 10-9 or 430 nm

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posted: | 5/9/2013 |

language: | English |

pages: | 23 |

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