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This problem on the topic of interference, we are shown in the figure waves along rays 1 and 2 that are initially in phase with the same wavelength lambda in a.
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However, ray 2 goes through a material with length l and index refraction n.
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The rays are then reflected by mirrors to a common point p that is on a screen, and we want to suppose that the length of l can vary from 0 to 2 ,400 nanometers.
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We want to suppose also that from l is equal to 0 to l is equal to 900 nanometers, the intensity i of the light at point p varies with l as we can see in the graph.
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We want to know the values of l greater than ls where the intensity is maximum and zero, and we want to know the multiple of lambda that will give the phase difference between the two rays at a common point p when l is equal to 1 ,200 nanometers.
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Now for completely destructive interference, the intensity produced by the two waves is zero.
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When the interference between the two waves is completely destructive, their phase difference phi is given by 2m plus 1 times pi, where m is equal to 0, 1, 2, etc, any positive integer.
01:23
The equivalent condition is their part length difference is an odd multiple of lambda over 2, where lambda is the wavelength of the light...