In a Young's double slit experiment, the intensity at some point on the screen is found to be times of the maximum

Question

In a Young's double slit experiment, the intensity at some point on the screen is found to be $\frac{3}{4}$ times of the maximum of the interference pattern. The path difference between the interfering waves at this point is $\frac{λ}{x}$ , where $λ$ is wavelength of the incident light. The value of $x$ is ______. [2026]

Smart Achievers · Accepted Answer

(6)

$I_{r} = I_{\max} \cos^{2} \frac{ϕ}{2}$

$Given: Intensity I = \frac{3}{4} I_{\max} [at any point]$

$\frac{3 I_{\max}}{4} = I_{\max} \cos^{2} \frac{ϕ}{2}$

$\cos \frac{ϕ}{2} = \frac{\sqrt{3}}{2}$

$\frac{ϕ}{2} = 30 ° = \frac{π}{6} \Rightarrow ϕ = \frac{π}{3}$

$Path difference Δ x = \frac{ϕ}{2 π} \times λ$

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

$Δ x = \frac{λ}{6} \Rightarrow x = 6$

Solution

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Solution

Ans. (6) Ir=Imaxcos2ϕ2 Given: Intensity I=34Imax [at any point] 3Imax4=Imaxcos2ϕ2 cosϕ2=32 ϕ2=30°=π6⇒ϕ=π3 Path difference Δx=ϕ2π×λ ⇒Δx=π3×2π×λ Δx=λ6⇒x=6

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