Optics Question 225
A parallel monochromatic beam of light is incident normally on a narrow slit. A diffraction pattern is formed on a screen placed perpendicular to the direction of incident beam. At the first maximum of the diffraction pattern the phase difference between the rays coming from the edges of the slit is π radians
[IIT?JEE 1995, 98]
Options:
0
B) $ \frac{\pi }{2} $
C) $ \pi $
D) $ 2\pi $
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Answer:
Correct Answer: D
Solution:
The phase difference $ (\varphi ) $ between the wavelets from the top edge and the bottom edge of the slit is $ \varphi =\frac{2\pi }{\lambda }(d\sin \theta ) $ where d is the slit separation.
The first minima of the diffraction pattern occurs at $ \sin \theta =\frac{\lambda }{d} $ so $ \varphi =\frac{2\pi }{d}( d\times \frac{\lambda }{d} )=\frac{2\pi \lambda }{d} $
BETA
