Optics Question 876
Question: The maximum number of possible interference maxima for slit separation equal to $ 1.8\lambda $ , where $ \lambda $ is the wavelength of light used, in a Young’s double slit experiment is
Options:
A) zero
B) 3
C) infinite
D) 5
Show Answer
Answer:
Correct Answer: B
Solution:
[b] As $ \sin \theta =\frac{n\lambda }{d} $ and $ \sin \theta $ cannot be $ 1 $
$ \therefore 1=\frac{n\lambda }{1.8\lambda } $ or $ n=1.8 $ .
Hence maximum number of possible interference maximas, 0, $ ~\pm 1 $ i.e. 3
BETA
