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In the arrangement shown wavelength of light used is $λ$ . The distance between slits $S_{1}$ and $S_{2}$ is d (<< D). $S_{4}$ is on the axis of symmetry as shown. The distance between $S_{3}$ and $S_{4}$ is $u = \frac{λ D}{3 d} .$ If the ratio of maximum to minimum intensity observed on screen is k. Find k.

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Detailed Solution

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If intensity of light incident on slits $S_{1}$ and $S_{2}$ is $I_{0}$ .
The intensity through $S_{4}$ will be $I_{1} = 4 I_{0}$
The intensity through $S_{3}$ Will be $I_{2} = 4 I_{0} \cos^{2} (\frac{ϕ}{2}) = I_{0}$
$ K = \frac{I_{\max}}{I_{\min}} = {(\frac{\sqrt{4 I_{0}} + \sqrt{I_{0}}}{\sqrt{4 I_{0}} - \sqrt{I_{0}}})}^{2} = {(\frac{3}{1})}^{2} = \frac{9}{1} .$

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In the arrangement shown wavelength of light used is λ . The distance between slits S1 and S2 is d (<< D). S4 is on the axis of symmetry as shown. The distance between S3 and S4 is u=λD3d. If the ratio of maximum to minimum intensity observed on screen is k. Find k.