  Question

In a Young's double slit experiment the intensity of the resultant wave at a point P on the screen is I where the path difference between the waves from coherent sources $$S_1$$ and $$S_2$$ is $$\lambda$$. Then the intensity of the resultant wave at a point where the path difference is $$\lambda /4$$ is given by.

A
I/2  B
2I  C
4I  D
I/2  Solution

The correct option is D $$I/2$$Given,The intensity of the resultant wave at a point P on the screen is $$I$$ when the path difference is $$\lambda$$Here we have to find the intensity of the resultant wave at a point where the path difference is $$\lambda/4$$.We have Intensity, $$I = I$$ $$\cos^2\dfrac\phi2$$         ............... (1)Where $$\phi$$ is phase differenceAlso, Phase difference, $$\phi$$  $$= \dfrac {2\pi}{\lambda}$$ X path difference      .................(2)Given, path difference is $$\dfrac{\lambda}{4}$$So, pahse difference, $$\phi$$ = $$\dfrac{2\pi}{\lambda}$$ X $$\dfrac\lambda4$$That is,                        $$\phi$$ = $$\dfrac\pi2$$ Substituting the value of $$\phi$$ in equation (1) we get,  Intensity, $$I$$ = $$I$$ $$\cos^2\dfrac\pi4$$                     Physics

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