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# What is the angular separation between two lines of $\lambda_1 =5882 A^{\circ}$ and $\lambda_2 =5852A^{\circ}$ in first order grating with 6000 lines per cm in normal incidence.

$(a)\;2.5 \times 10^{-3}\;rad \\ (b)\;1.8 \times 10^{-3}\;rad \\ (c)\;3.6 \times 10^{-3}\;rad \\ (d)\;1.8 \times 10^{-4}\;rad$

For normal incidence
$n \lambda = d\sin \theta$
$n=1$
$\sin \theta =\large\frac{\lambda}{d}$
$(\sin \theta_1 - \sin \theta _2 ) =\large\frac{1}{d} $$(\lambda_1 -\lambda_2) For small angle \theta_1 -\theta_2 =\large\frac{1}{d}$$ (5882 -5852) \times 10^{-10}$
Since there are 6000 lines in 1 cm
$d= \large\frac{10^{-2}}{6000}$$m$
$\therefore (\theta_1 -\theta_2 )= \large\frac{30 \times 10^{-10}}{\Large\frac{10^{-2}}{6000}}$
$\qquad= 1.8 \times 10^{-3} rad$
Hence b is the correct answer.