Neet Solved Paper 2015 Question 37
Question: For a parallel beam of monochromatic light of wavelength ’ $ \lambda $ ’ diffraction is produced by a single slit whose width ‘a’ is of the order of the wavelength of the light. If D is the distance of the screen from the slit, the width a the of the central maxima will be
Options:
A) $ \frac{2\lambda D}{a} $
B) $ \frac{D\lambda }{a} $
C) $ \frac{Da}{\lambda } $
D) $ \frac{2Da}{\lambda } $
Show Answer
Answer:
Correct Answer: A
Solution:
For the condition of maxima $ \sin \theta =\frac{\lambda }{a} $
From the geometry, $ \sin \theta =\theta =\frac{Y}{D} $ (for small angle) So, $ \frac{Y}{D}=\frac{\lambda }{a} $
Þ $ ,Y=\frac{\lambda D}{a} $ Hence, width of central maxima = $ 2Y=\frac{2\lambda D}{a} $
BETA
