Download Interference And Diffraction-Classical Physics-Handouts and more Lecture notes Classical Physics in PDF only on Docsity! PHYSICS –PHY101 VU © Copyright Virtual University of Pakistan 102 destructive constructive Summary of Lecture 33 – INTERFERENCE AND DIFFRACTION 1. Two waves (of any kind) add up together, with the net result being the simple sum of the two waves. Consider two waves, both of the same frequency, shown below. If they start together (i.e. are with each other) then the net amplitude is increased. This is called . But if they start at different times (i.e. are with each other) th in phase constructive interference out of phase en the net amplitude is decreased. This is called .destructive interference In the example above, both waves have the same frequency and amplitude, and so the resulting amplitude is doubled (constructive) or zero (destructive). But interference occurs for any two waves even when their amplitudes and frequencies are different. 2. Although any waves from different sources interfere, if one wants to observe the interference of light then it is necessary to have a source of light. Coherent means that both coherent waves should have a fixed phase relative to each other. Even with lasers, it is very difficult to produce coherent light from two separate sources. Observing interference usually requires taking two waves from a single source, with each going along a different path. In the figure, an incoherent light source illuminates the first slit. This creates a uniform and coherent ill 1 2 umination of the second screen. Then waves from the slits S and S meet on the third screen and create a pattern of alternating light and dark fringes. 3. Wherever there is a bright fringe, constructive interference has occured, and wherever there is a dark fringe, destructive interference has occured. We shall now calculate where on the third 1 2 screen the interference is constructive. Take any point on the third screen. Light reaches this point from both S and S , but it will take different amounts of time to get there. Hence there will be a phase difference that we can calculate. Look at the diagram below. You can see that light from one of the slits has to travel an extra docsity.com PHYSICS –PHY101 VU © Copyright Virtual University of Pakistan 103 distance equal to sin , and so the extra amount of time it takes is ( sin ) / . There will be constructive interference if this is equal to ,2 , 3 , (remember that the time period is d d c T T T θ θ ⋅ ⋅ ⋅ inversely related to the frequency, 1/ , and that ). sin We find that sin , where 1,2,3, What about for destructive interference? Here the waves will T c d nnT d n c n ν λν θ θ λ ν = = = = ⇒ = = ⋅ ⋅ ⋅ cancel each other 3 5 if the extra amount of time is , , , The 2 2 2 1 condition then becomes sin ( ) . 2 T T T d nθ λ ⋅ ⋅ ⋅ = + -54. Example: two slits with a separation of 8.5× 10 m create an interference pattern on a screen 2.3m away. If the 10 bright fringe above the central is a linear distance of 12 cm from it, n = 1 1 what is the wavelength of light used in the experiment? Answer: First calculate the angle to the tenth bright fringe using tan . Solving for gives, 0.12 tan tan 2.3 y L y m L m θ θ θ − − = ⎛ ⎞ ⎛ ⎞= =⎜ ⎟ ⎜ ⎟ ⎝ ⎠ ⎝ ⎠ 3.0 . From this,= 5 78.5 10 sin sin(3.0 ) 4.4 10 440 (nanometres). 10 d m nm n λ θ − −⎛ ⎞×= = = × =⎜ ⎟ ⎝ ⎠ 5. When a wave is reflected at the interface of two media, the phase will not change if it goes from larger refractive index to a smaller one. But for smaller to larger, there will be phase change of a half-wavelength. One can show this using Maxwell's equations and applying the boundary conditions, but this will require some more advanced studies. Instead let's just use this fact below. docsity.com
