A parallel beam of wavelength $λ = 450 \times 10^{- 9} m$ passes through a long slit of width $2 \times 10^{- 4} m$ . The angular divergence for which most of light is diffracted is

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$4.5 \times 10^{- 3} rad$

$3.5 \times 10^{- 3} rad$

$5.4 \times 10^{- 4} rad$

$2.8 \times 10^{- 4} rad$

answer is B.

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Detailed Solution

Most of the light is diffracted between the two first order minima.

$∴$ Required angle $= 2 θ$

$= 2 sinθ = \frac{2 λ}{d} = \frac{2 \times 450 \times 10^{- 9}}{2 \times 10^{- 4}}$

$= 450 \times 10^{- 5} radian = 4 .5 \times 10^{- 3} radian$

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