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physics help p2

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Bhai will anyone explain me jun 09 ques 5 Variant 1. Its really tough yar....
 
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well, this question was a pain!

since s1 and s2 are coherent, and they're asking us to find the minima, which means odd multiples of π
so, you take phase difference = (2n+1)π for the minima okay, and then we know that phase difference is also equal to phase difference= (2π/λ)x , where x is the path difference and the path difference we can calculate from the diagram, by taking the hypotenuse from S2 and M, we get 128, then 128-100, coz the path difference from M then we get 28.
equate these 2 formulas (2π/λ)x=(2n+1)π - we get λ by using v=fλ then, λ-v/f - where f is 4kHz-1kHz which is 3kHz okay, then

then, π gets cancelled from both the sides,

(2/(330/3000))x28/100 = (2n+1) --- 28/100 coz its in cms we need it in meters
(2/0.11)x28/100=(2n+1)
n=2.045454545
therefore n = two minma!

Over and Out
Jurol
 
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