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(A) Let $I_{0}$ be the intensity of each individual slit. The resultant intensity $I$ at any point is given by $I = 4I_{0} \cos^{2}(\phi/2)$,where $\p

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$K/4$

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(A) Let $I_{0}$ be the intensity of each individual slit. The resultant intensity $I$ at any point is given by $I = 4I_{0} \cos^{2}(\phi/2)$,where $\phi$ is the phase difference.Phase difference $\phi = \frac{2\pi}{\lambda} 	imes (	ext{path difference})$.For path difference $\Delta x = \lambda$,the phase difference $\phi = \frac{2\pi}{\lambda} 	imes \lambda = 2\pi$.The intensity $I = 4I_{0} \cos^{2}(2\pi/2) = 4I_{0} \cos^{2}(\pi) = 4I_{0}(1)^{2} = 4I_{0}$.Given that this intensity is $K$,we have $4I_{0} = K$,so $I_{0} = K/4$.Now,for path difference $\Delta x = \lambda/3$,the phase difference $\phi = \frac{2\pi}{\lambda} 	imes \frac{\lambda}{3} = \frac{2\pi}{3}$.The new intensity $I' = 4I_{0} \cos^{2}(\phi/2) = 4I_{0} \cos^{2}(\frac{2\pi/3}{2}) = 4I_{0} \cos^{2}(\pi/3)$.Since $\cos(\pi/3) = 1/2$,we have $I' = 4I_{0} 	imes (1/2)^{2} = 4I_{0} 	imes (1/4) = I_{0}$.Substituting $I_{0} = K/4$,we get $I' = K/4$ units.

In Young's double-slit experiment using monochromatic light of wavelength $\lambda$,the intensity of light at a point on the screen where the path difference is $\lambda$ is $K$ units. What is the intensity of light at a point where the path difference is $\lambda / 3$?

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