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(D) In Young's double slit experiment,the intensity at any point on the screen is the sum of the intensities due to individual wavelengths because the

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$8 I_{0}$

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(D) In Young's double slit experiment,the intensity at any point on the screen is the sum of the intensities due to individual wavelengths because they are incoherent with respect to each other.For a single wavelength with slit intensity $I_{0}$,the maximum intensity is given by $I_{\max} = (\sqrt{I_{0}} + \sqrt{I_{0}})^2 = 4I_{0}$.Since the source emits two wavelengths,$400 \,nm$ and $800 \,nm$,each with slit intensity $I_{0}$,the maximum intensity for each wavelength is $4I_{0}$.At the central maximum,both wavelengths produce their respective maximum intensities simultaneously.Therefore,the total maximum intensity is $I_{	ext{total}} = 4I_{0} + 4I_{0} = 8I_{0}$.

$A$ source simultaneously emitting light at two wavelengths $400 \,nm$ and $800 \,nm$ is used in the Young's double slit experiment. If the intensity of light at the slit for each wavelength is $I_{0}$,then the maximum intensity that can be observed at any point on the screen is

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