Optics Question 901
Question: A parallel beam of light $ (\lambda =5000\overset{o}{\mathop{A}}) $ is incident at an angle $ \alpha =30{}^\circ $ with the normal to the slit plane in YDSE. Assume that the intensity due to each slit at any point on the screen is $ I _{0} $ . Point O is equidistant from $ S _{1} $ and $ S _{2} $ . The distance between slit is 1 mm, then the intensity at
Options:
A) O is $ 3I _{0} $
B) O is zero
C) a point 1 m below O is $ 4I _{0} $
D) a point on the screen 1 m below O is zero
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Answer:
Correct Answer: C
Solution:
[c] The path difference at O, $ \Delta x=d\sin \alpha =d\sin 30^{o}=\frac{{{10}^{-3}}}{2}m $
Now $ \phi =\frac{2\pi }{\lambda }.\Delta x=\frac{2\pi }{5000\times {{10}^{-10}}}\times \frac{{{10}^{-3}}}{2}=2\pi \times 10^{3} $
So $ I=I _{o}+I _{o}+2\sqrt{I _{o}I _{o}}\cos (2\pi \times 10^{3})=4I _{o} $
The angular position of P, $ \tan \theta =\frac{1}{\sqrt{3}} $ ; or $ \theta =30^{o} $ .
It means path $ diff^{n} $ at P is also $ \Delta x $
and hence $ I=4I _{o} $
BETA
