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# the path difference between two interfeering waves ata point on the screen is x/3, where x is wave leength , from the central maximum, the ratio of intensity at this point and at the central fringe will be

Grade:12th pass

## 1 Answers

Neeti
571 Points
5 years ago
$\Delta$x/$\lambda$ = $\Delta$$\Phi$/2$\pi$ so we have path difference delta x = x/3 where  x= $\lambda$ = wavelength.
From here we get phase difference between the two =$\Delta$ $\Phi$
Now when phase difference is $\Phi$ the resultant intensity can be calculated by :
Inet =I1+ I2 + sq.root (I1xI2)cos $\Phi$ (here we assume I1 = I2 = I ) and intensity at central maxima = 4I (since $\Delta$$\Phi$ = zero for center, Inet = 4I )

So you get the ratio. ( Inet / 4I )

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