Optics Question 142
Question: In an interference arrangement similar to Young’s double slit experiment, the slits S1 and S2 are illuminated with coherent microwave sources each of frequency 106 Hz. The sources are synchronized to have zero phase difference. The slits are separated by distance d = 150 m. The intensity I $ (\theta ) $ is measured as a function ofq, where q is defined as shown. If I0 is maximum intensity, then $ I(\theta ) $ for $ 0\le \theta \le 90^{o} $ is given by
[IIT 1995]
Options:
A) $ I(\theta )=I _{0} $ for $ \theta =0^{o} $
B) $ I(\theta )=I _{0}/2 $ for $ \theta =30^{o} $
C) $ I(\theta )=I _{0}/4 $ for $ \theta =90^{o} $
D) $ I(\theta ) $ is constant for all values of q
Show Answer
Answer:
Correct Answer: B
Solution:
For microwave $ \lambda =\frac{c}{f}=\frac{3\times 10^{8}}{10^{6}}=300\ m $
As $ \Delta x=d\sin \theta $
Phase difference $ \varphi =\frac{2\pi }{\lambda } $
(Path difference) $ =\frac{2\pi }{\lambda }(d\sin \theta )=\frac{2\pi }{300}(150\sin \theta )=\pi \sin \theta $
$ I _{R}=I _{1}+I _{2}+2\sqrt{I _{1}I _{2}}\cos \varphi $
Here $ I _{1}=I _{2} $ and $ \varphi =\pi \sin \theta $
$ \therefore I _{R}=2I _{1}[1+\cos (\pi \sin \theta )]=4I _{1}{{\cos }^{2}}( \frac{\pi \sin \theta }{2} ) $
IR will be maximum when $ {{\cos }^{2}}( \frac{\pi \sin \theta }{2} )=1 $
$ \therefore {{(I _{R})} _{\max }}=4I _{1}=I _{o} $
Hence $ I=I _{o}{{\cos }^{2}}( \frac{\pi \sin \theta }{2} ) $
If $ \theta =0, $ then $ I=I _{o}\cos \theta =I _{o} $
If $ \theta =30{}^\circ , $
then $ I=I _{o}{{\cos }^{2}}(\pi /4)=I _{o}/2 $
If $ \theta =90^{o}, $ then $ I=I _{o}{{\cos }^{2}}(\pi /2)=0 $
BETA
