Questions

In a Young’s double slit experiment $d = 1 m m, λ = 6000 A^{0} and D = 1 m$ (where d, $λ$ and D have usual meaning). Each of slit individually produces the same intensity on the screen. Then select the correct option(s).

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The minimum distance between two points on the screen having 75% intensity of the maximum intensity is 0.2 mm

The minimum distance between two points on the screen having 75% intensity of the maximum intensity is 20 μm

Fringe width is 0.6 mm

Fringe width is 0.3 mm

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detailed solution

Correct option is A

β=Dλd=0.6 mmPhase difference between centre and any point is ϕ Then I=4I0cos2ϕ2=34(4I0) ⇒cosϕ2=±32So all point at distance 60o360oβ=β6 from a local maximum will have 34th of maximum intensity.∴ minimum distance between such points =β3 = 0.2 mm

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In an interference arrangement, similar to Young’s double slit experiment, the slits S₁ and S₂ are illuminated with coherent microwave sources, each of frequency 106 Hz. The sources are synchronized to have zero phase difference. The slits are separated by distance d = 150.0 m. The intensity $I_{(θ)}$ is measured as a function of $θ$ , where $θ$ is defined as shown in the figure. If I₀ is maximum intensity, then $I_{(θ)}$ for
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