Suppose that one of the slits in YDSE is wider than the other, so the amplitude of the light reaching on screen from one slit, acting alone, is thrice that from the other slit, acting alone. The intensity for Central Bright Fringe is $I_{0} = 5 W / m^{2}$ . Then find the intensity $I (W / m^{2})$ on the screen for (angular position) $θ = 37 °$ . (Take $d / λ = 5 / 6,$ where $d$ is the separation between the slits and $λ$ is the Wavelength of the light)

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Detailed Solution

$|_{C . B . F .} = K [A_{1}^{2} + A_{2}^{2} + 2 A_{1} A_{2} \cos ϕ] = |_{0} K A^{2} = \frac{|_{0}}{16}, (A_{1} = A, A_{2} = 3 A a n d ϕ = 0 °) |_{θ} = \frac{16}{25} K [A_{1}^{2} + A_{2}^{2} + 2 A_{1} A_{2} \cos ϕ] = 1 ϕ = \frac{2 π}{λ} Δ x = \frac{2 π}{λ} \times d \sin θ = \frac{2 π}{λ} \times d \frac{3}{5} S o, ϕ = π$