In a Young's double slit experiment, the intensity at a point is of the maximum intensity

Question

In a Young's double slit experiment, the intensity at a point is ${(\frac{1}{4})}^{th}$ of the maximum intensity, the minimum distance of the point from the central maximum is __________ $μ m$ . $(Given λ = 600 nm, d = 1.0 mm, D = 1.0 m)$

Smart Achievers · Accepted Answer

(200)

$Intensity at a point is given by$

$I = I_{0} \cos^{2} (\frac{Δ ϕ}{2})$

$\Rightarrow According to question \frac{I_{0}}{4} = \cos^{2} (\frac{Δ ϕ}{2})$

$Δ ϕ = \frac{2 π}{3}$

$∴$ $Δ ϕ = \frac{2 π}{λ} (\frac{y d}{D}) = \frac{2 π}{3}$

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

Solution

Q.
In a Young's double slit experiment, the intensity at a point is ${(\frac{1}{4})}^{th}$ of the maximum intensity, the minimum distance of the point from the central maximum is __________ $μ m$ . $(Given λ = 600 nm, d = 1.0 mm, D = 1.0 m)$

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Solution

Q. In a Young's double slit experiment, the intensity at a point is (14)th of the maximum intensity, the minimum distance of the point from the central maximum is __________ μm. (Given λ=600 nm, d=1.0 mm, D=1.0 m)

Ans. (200) Intensity at a point is given by I=I0cos2(Δϕ2) ⇒According to question I04=cos2(Δϕ2) Δϕ=2π3 ∴ Δϕ=2πλ(ydD)=2π3 ⇒y=λD3d=600×10-9×13×10-3=2×10-4 m

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Q.
In a Young's double slit experiment, the intensity at a point is ${(\frac{1}{4})}^{th}$ of the maximum intensity, the minimum distance of the point from the central maximum is __________ $μ m$ . $(Given λ = 600 nm, d = 1.0 mm, D = 1.0 m)$