$(a)\;60^{\circ} \\ (b)\;30^{\circ} \\ (c)\;15^{\circ} \\ (d)\;0^{\circ} $

For diffraction,

For central maxima

$a \sin \theta =\lambda$

$\sin \theta =\large\frac{\lambda}{a}$

$\theta= \sin ^{-1} \bigg( \large\frac{\lambda}{a}\bigg)$

here $\lambda = 2\times 10^{-3}$

$a= 0.04 m$

$\quad= 4 \times 10^{-3}\;m$

$\theta =\sin ^{-1} \large\frac{1}{2}$

$\qquad= 30^{\circ}$

Hence b is the correct answer.

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