Two light waves of wavelengths $800 nm$ and $600 nm$ are used in Young’s double slit experiment to obtain interference fringes on a screen placed $7 m$ away from plane of slits. If the two slits are separated by 0.35 mm, then shortest distance from the central bright maximum to the point where the bright fringes of the two wavelength coincide will be ______mm.

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answer is 48.

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Detailed Solution

$\begin{array}{l} d = 0.35 mm = 35 \times 10^{- 5} m \\ D = 7 m \Rightarrow y = n_{1} β_{1} = n_{2} β_{2} \\ y = n_{1} \frac{λ_{1} D}{d} = n_{2} \frac{λ_{2} D}{d} \\ n_{1} λ_{1} = n_{2} λ_{2} \Rightarrow \frac{n_{1}}{n_{2}} = \frac{λ_{2}}{λ_{1}} = \frac{600}{800} = \frac{3}{4} \\ ∴ y = \frac{3 λ_{1} D}{d} = \frac{4 λ_{2} D}{d} \\ y = \frac{3 \times 800 \times 10^{- 9} \times 7}{35 \times 10^{- 5}} = 4.8 \times 10^{- 2} m \\ = 48 \times 10^{- 3} m \\ y = 48 mm \end{array}$

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