In Young's double slit experiment , the light has a frequency of `6 xx 10^(14) Hz` and the distance between the centres of adjacent fringes is `0.75 mm`. If the screen is `1.5m` away, what is the distance between the slits ?
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Here, `v = 6 xx 10^(14) Hz`, `beta = 0.75 xx 10^(-3)m, D = 1.5m, d = ?` `lambda = ( c)/(v) = (3 xx 10^(8))/(6 xx 10^(14)) = 5 xx 10^(-7) m` As `beta = (lambda D)/(d), d = (lambda D)/(beta) = (5 xx 10^(-7) xx 1.5)/(0.75 xx 10^(-3))` `= 10^(-3)m`
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