A) \[6\,mm\]
B) \[12\,mm\]
C) \[3\,mm\]
D) \[9\,mm\]
Correct Answer: A
Solution :
For secondary maxima \[d\sin \theta =\frac{5\lambda }{2}\] \[\Rightarrow d\theta \,=d.\frac{x}{D(\approx \,f)}=\frac{5\lambda }{2}\] \[\Rightarrow 2x=\frac{5\lambda f}{d}=\frac{5\times 0.8\times {{10}^{-7}}}{4\times {{10}^{-4}}}=6\times {{10}^{-3}}m=6mm\]
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