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Light is incident normally on a diffraction grating through first order diffraction is seen at $32^{\circ}$ second order diffraction will be seen at

$(a)\;16^{\circ} \\ (b)\;62^{\circ} \\ (c)\;32^{\circ} \\ (d)\;no\;second\; order\; diffraction \;takes \;place $

1 Answer

1st order diffraction
$d \sin \theta =n \lambda$
$\therefore d \sin \theta = \lambda$
$ d \sin 32^{\circ}=\lambda$
For second order
$d \sin \theta_2=2 \lambda$
$\qquad= 2 d \sin 32^{\circ}$
$\sin \theta_2= 2 \sin 32^{\circ}$
$ \sin \theta_2 > 1$
Which is not possible
$\therefore$ no second order diffraction takes places.
Hence d is the correct answer.
answered Jan 23, 2014 by meena.p

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