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| Numerical - 20 |
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| 20. In Young's double slit experiment, the intensity at a point is 75% of maximum intensity. What is the smallest distance of this point from the central fringe? Given d = 0.1mm, D = 1m and l = 600 nm. |
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| Suggested solution: |
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| The intensity I at a point on the screen where the two disturbances arrive having a phase difference s is |
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| I = 0.751max Therefore, |
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| Let x be the minimum distance of this point from the central fringe. |
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| Then, |
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| or |
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