Light from a source emitting two wavelengths $λ_{1}$ and $λ_{2}$ is allowed to fall on Young’s double slit apparatus after filtering one of the wavelengths. The position of interference maxima is noted. When the filter is removed i.e. both the wavelengths are now incident on the apparatus, it is found that maximum intensity is now produced where the fourth maximum occurred previously. If the other wavelength is filtered at the same location the third maxima is found. If $λ_{1} : λ_{2}$ is 6:n, find n

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answer is 8.

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Detailed Solution

$N \frac{4 λ_{1} D}{d} = \frac{3 λ_{2} D}{d}, \frac{λ_{1}}{λ_{2}} = \frac{3}{4} = \frac{6}{8}$

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