Can any one do this optics problem?

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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)

#Physics #Optics

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3 Answers

rocky ·2010-02-15 23:17:25

is it A ????????

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Manmay kumar Mohanty ·2010-02-15 23:20:44

Yes it is A. Please explain.

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var ·2010-03-17 01:12:05

max intensity=I_ocos²(âˆ‚)/2 .... âˆ‚-phase diff..
given
max intensity=3I_o/4
so âˆ‚=60Â°

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

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Can any one do this optics problem?

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