Optics Question 188

Question: The maximum number of possible interference maxima for slit-separation equal to twice the wavelength in Young’s double-slit experiment is

[AIEEE 2004]

Options:

A) Infinite

B) Five

C) Three

D) Zero

Show Answer

Answer:

Correct Answer: B

Solution:

For maxima Δ=dsinθ=nλ

Therefore 2λsinθ=nλ

Therefore sinθ=n2 since value of sin q can not be greater 1. \ n = 0, 1, 2

Therefore only five maximas can be obtained on both side of the screen.



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