A) 4.5 cm
B) 1.5 cm
C) 6.0 cm
D) 3.0 cm
Correct Answer: D
Solution :
Min. \[f\sin \theta =n\lambda \] \[\sin \theta =\frac{n\lambda }{6}\] n = 2 \[\sin \theta =\frac{n\lambda }{6}=\tan {{\theta }_{1}}=\frac{{{x}_{1}}}{D}\] x = 4 \[\sin {{\theta }_{2}}=\frac{4\lambda }{6}=\frac{{{x}_{2}}}{D}\] \[{{x}_{2}}-{{x}_{1}}=\frac{4\lambda }{6}-\frac{2\lambda }{6}=\frac{2\lambda }{6}=3cm\] width of central \[\max =\frac{2\lambda }{6}=3cm\]You need to login to perform this action.
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