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



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