Can any one do this optics problem?

In Young's double slit exper., slits are horizontal. The intensity at a point P shown in fig. is 34I0, where I0, is the max. intensity. Then the value of \theta is,
(Given the distance between the two slits S1 and S2 is 2\lambda)

A) cos-1(112) B) sin -1 (112) C) tan -1 (112) D) sin -1 (35)

3 Answers

3
rocky ·

is it A ????????

1
Manmay kumar Mohanty ·

Yes it is A. Please explain.

1
var ·

max intensity=Iocos2(∂)/2 .... ∂-phase diff..
given
max intensity=3Io/4
so ∂=60°

draw perpendicular from S1P to S2P
..u ll get d path diff as 2Lambdacos(theta)..
path diff=(lambda)/2pie * ∂......
equate
nd u ll get d ans as (A)...

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