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Sahay Sir > Question Answers > Uncategorised (JEE Advanced Physics by BM Sharma + GMP Solutions) > In Young’s double slit experiment lemda = 500 nm, d = 1mm, and D = 4 m. The minimum distance from the central maximum for which the intensity is half of the maximum intensity

In Young’s double slit experiment lemda = 500 nm, d = 1mm, and D = 4 m. The minimum distance from the central maximum for which the intensity is half of the maximum intensity

and D = 4 m. The minimum distance from the central maximum for which the intensity is half of the maximum intensity d = 1mm In Young's double slit experiment lemda = 500 nm