A) 1 : 4 : 9
B) 1 : 2 : 3
C) \[1\,:\frac{4}{9\,{{\pi }^{2}}}:\frac{4}{25\,{{\pi }^{2}}}\]
D) \[1:\frac{1}{{{\pi }^{2}}}:\frac{9}{{{\pi }^{2}}}\]
Correct Answer: C
Solution :
\[I={{I}_{0}}{{\left[ \frac{\sin \alpha }{\alpha } \right]}^{2}},\] where \[\alpha =\frac{\varphi }{2}\] For \[{{n}^{th}}\] secondary maxima \[d\sin \theta =\left( \frac{2n+1}{2} \right)\lambda \] \[\Rightarrow \alpha =\frac{\varphi }{2}=\frac{\pi }{\lambda }\left[ d\sin \theta \right]=\left( \frac{2n+1}{2} \right)\pi \] \[\therefore \,\,\,I={{I}_{0}}{{\left[ \frac{\sin \left( \frac{2n+1}{2} \right)\pi }{\left( \frac{2n+1}{n} \right)\pi } \right]}^{2}}=\frac{{{I}_{0}}}{{{\left\{ \frac{(2n+1)}{2}\pi \right\}}^{2}}}\] So \[{{I}_{0}}:{{I}_{1}}:{{I}_{2}}={{I}_{0}}:\frac{4}{9{{\pi }^{2}}}{{I}_{0}}:\frac{4}{25{{\pi }^{2}}}{{I}_{0}}\] \[=1:\frac{4}{9{{\pi }^{2}}}:\frac{4}{25{{\pi }^{2}}}\]
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