1.

A screen is placed `2m` away from the single narrow slit. Calculate the slit width if the first minimum lies `5 mm` on either side of the central maximum. Incident plane waves have a wavelength of `5000 Å`.

Answer» Here, distance of the screen from the
slit, `D = 2m, a = ?, x = 5mm = 5 xx 10^(-3)m`,
`lambda = 5000 Å = 5000 xx 10^(-10)m`
For the first secondary minima, ,brgt `sin theta = (lambda)/(a) = (x)/(D)`
`:. a = (D lambda)/(x) = (2 xx 5000 xx 10^(-10))/(5 xx 10^(-3)) = 2 xx 10^(-4)m`


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