Light from a monochromatic point source S is incident on the two slits of an opaque wall placed at a distance L as shown in the diagram, Intensity of interference pattern is observed at the point P on the screen placed at a distance L behind the slits. When the upper slit is shifted upward then the distance between the slits is changed from $200 μ m$ to $400 μ m$ , the intensity at P changes from $I_{\max}$ to $\frac{3}{4} I_{\max}$ . If L=1 m, then find the wave length of light used $(i n A)$

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answer is 4800.

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Detailed Solution

$\frac{3}{4} |_{\max} = |_{\max} \cos^{2} \frac{ϕ}{2} \Rightarrow \cos \frac{ϕ}{2} = \frac{\sqrt{3}}{2} \Rightarrow \frac{ϕ}{2} = \frac{π}{6} \Rightarrow ϕ = π / 3 ϕ = \frac{2 π}{λ} Δ x = \frac{π}{3} \Rightarrow Δ x = \frac{λ}{6} G i v e n d = 200 μ m = 2 \times 10^{- 4} m$

$2 [\sqrt{L^{2} + \frac{9 d^{2}}{4}} - \sqrt{L^{2} + \frac{d^{2}}{4}}] = \frac{λ}{6} \Rightarrow \frac{2 d^{2}}{L} = \frac{λ}{6} \Rightarrow λ = \frac{12 d^{2}}{L} = \frac{12 \times 4 \times 10^{- 8}}{1} = 4800 A$