15. How many interference fringes will be seen the separation
the size of each slitis (1/5)th the separation between the
two sits
a) n. 10 b) na 5 c) n.15 d) 2
Answers
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6
Explanation:
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Answered by
0
Answer:
10
Explanation:
n x lambda x D / a = 2 x lambda x D / a ---------> 1.
where,
n is the number of fringes
lambda is the wavelength
D is the distance between the source slits and the screen.
d is the distance between the slits
a is the size of the slits
Given that a=d/5
substituting in eq. 1
n x lambda x D / d/5 = 2 x lambda x D / d/5
on simplifying,
n=2
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