$(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 $

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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.

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