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A screen is placed `50 cm` from a single slit, which is illuminated with `6000 Å` light. If the distance between the first and third minima in the diffraction pattern is `3.00 mm`, what is the width of the slit ?

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Correct Answer - `0.2 mm`
Using `x = (lambda D)/(d)`
`x_(2) - x_(1) = (3 lambda - lambda)(D)/(d) = (2 lambda D)/(d)`
`d = (2 lambda D)/(x_(2) - x_(1)) = (2 xx 6000 xx 10^(-10) xx 0.5)/(3 xx 10^(-3))`
`d = 2 xx 10^(-4)m = 0.2 mm`

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