A) 6 mm
B) 12 mm
C) 3 mm
D) 9 mm
Correct Answer: A
Solution :
[a] 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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