In Y.D.S.E. using light of wavelength lambda, | vedclass

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(B) Let the intensity of each individual slit be $I_0$. The resultant intensity is given by $I_R = 4I_0 \cos^2(\frac{\Delta \phi}{2})$.For a path diff

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$\frac{M}{4}$

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(B) Let the intensity of each individual slit be $I_0$. The resultant intensity is given by $I_R = 4I_0 \cos^2(\frac{\Delta \phi}{2})$.For a path difference $\Delta x = \lambda$,the phase difference $\Delta \phi = \frac{2\pi}{\lambda} \cdot \lambda = 2\pi$.Given $I_R = M$ at $\Delta x = \lambda$,we have $M = 4I_0 \cos^2(\pi) = 4I_0(1)^2 = 4I_0$. Thus,$I_0 = \frac{M}{4}$.For a path difference $\Delta x = \frac{\lambda}{3}$,the phase difference $\Delta \phi = \frac{2\pi}{\lambda} \cdot \frac{\lambda}{3} = \frac{2\pi}{3}$.The resultant intensity is $I_R' = 4I_0 \cos^2(\frac{2\pi/3}{2}) = 4I_0 \cos^2(\frac{\pi}{3})$.Substituting $I_0 = \frac{M}{4}$ and $\cos(\frac{\pi}{3}) = \frac{1}{2}$,we get $I_R' = 4(\frac{M}{4}) \cdot (\frac{1}{2})^2 = M \cdot \frac{1}{4} = \frac{M}{4}$.

In $Y.D.S.E.$ using light of wavelength $\lambda$,the intensity of light at a point on the screen with path difference $\lambda$ is $M$ units. Calculate the intensity of light at a point where the path difference is $\frac{\lambda}{3}$.

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