Optics Question 170
Question: The ratio of intensities of consecutive maxima in the diffraction pattern due to a single slit is
Options:
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}}} $
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Answer:
Correct Answer: C
Solution:
$ I=I _{0}{{[ \frac{\sin \alpha }{\alpha } ]}^{2}}, $ where $ \alpha =\frac{\varphi }{2} $ .
For $ n^{th} $ secondary maxima $ d\sin \theta =( \frac{2n+1}{2} )\lambda $
$ \Rightarrow \alpha =\frac{\varphi }{2}=\frac{\pi }{\lambda }[ d\sin \theta ]=( \frac{2n+1}{2} )\pi $
$ \therefore I=I _{0}{{[ \frac{\sin ( \frac{2n+1}{2} )\pi }{( \frac{2n+1}{n} )\pi } ]}^{2}}=\frac{I _{0}}{{{{ \frac{(2n+1)}{2}\pi }}^{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}}} $
BETA
