Physics And Measurement Question 298
Question: Intensity observed in an interference pattern is $ I=I _{0}Sin^{2} $ . $ At\theta =30^{o} $ intensity $ I=5\pm 0.0020W/m^{2} $ . Find percentage error in angle if $ I _{0}=20W/m^{2} $
Options:
A) $ \frac{4}{\pi }\sqrt{3}\times {{10}^{-2}}% $
B) $ \frac{2}{\pi }\sqrt{3}\times {{10}^{-2}}% $
C) $ \frac{1}{\pi }\sqrt{3}\times {{10}^{-2}}% $
D) $ \frac{3}{\pi }\sqrt{3}\times {{10}^{-2}}% $
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Answer:
Correct Answer: A
Solution:
[a] $ I=I _{0}{{\sin }^{2}}\theta \Rightarrow \theta ={{\sin }^{-1}}\sqrt{\frac{I}{I _{0}}} $
$ d\theta =\frac{1}{2}\frac{\sqrt{I _{0}}}{\sqrt{I _{0}-I}}\frac{dI}{\sqrt{I _{0}\sqrt{I}}} $
$ \frac{d\theta }{\theta }=\frac{1}{2}\frac{dI}{\sqrt{I(I _{0}-I)}{{\sin }^{-1}}\sqrt{I/I _{0}}} $
$ \therefore %error=\frac{4}{\pi }\sqrt{3}\times {{10}^{-2}}% $
