In a Young’s double slit experiment, the intensity at a point is ${(\frac{1}{4})}^{t h}$ of the maximum intensity. The minimum distance of the point from the central maximum is _____ µm.

(Given: $λ = 600 n m,$ $d = 1.0 m m,$ $D = 1.0 m$ ) [2024]

Question

In a Young’s double slit experiment, the intensity at a point is ${(\frac{1}{4})}^{t h}$ of the maximum intensity. The minimum distance of the point from the central maximum is _____ µm.

(Given: $λ = 600 n m,$ $d = 1.0 m m,$ $D = 1.0 m$ ) [2024]

Smart Achievers · Accepted Answer

(200) $I = I_{0} \cos^{2} (\frac{Δ ϕ}{2})$

$\frac{1}{4} = \cos^{2} (\frac{Δ ϕ}{2})$

$Δ ϕ = \frac{2 π}{λ} Δ x \Rightarrow Δ ϕ = \frac{2 π}{3} = \frac{2 π}{λ} Δ x \Rightarrow Δ x = \frac{λ}{3}$

$\Rightarrow Δ x = \frac{d y}{D} = \frac{λ}{3}$

$\Rightarrow y = \frac{λ D}{3 d} = \frac{600 \times 10^{- 9} \times 1}{3 \times 10^{- 3}} = 2 \times 10^{- 4} m = 200 μ m$

Solution

Q.
In a Young’s double slit experiment, the intensity at a point is ${(\frac{1}{4})}^{t h}$ of the maximum intensity. The minimum distance of the point from the central maximum is _____ µm.

(Given: $λ = 600 n m,$ $d = 1.0 m m,$ $D = 1.0 m$ ) [2024]

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Solution

Q. In a Young’s double slit experiment, the intensity at a point is (14)thof the maximum intensity. The minimum distance of the point from the central maximum is _____ µm. (Given: λ=600nm, d=1.0mm, D=1.0m) [2024]

Ans. (200) I=I0cos2(Δϕ2) 14=cos2(Δϕ2) Δϕ=2πλΔx⇒Δϕ=2π3=2πλΔx⇒Δx=λ3 ⇒Δx=dyD=λ3 ⇒y=λD3d=600×10-9×13×10-3=2×10-4m=200μm

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Q.
In a Young’s double slit experiment, the intensity at a point is ${(\frac{1}{4})}^{t h}$ of the maximum intensity. The minimum distance of the point from the central maximum is _____ µm.

(Given: $λ = 600 n m,$ $d = 1.0 m m,$ $D = 1.0 m$ ) [2024]