The intensity of each of the two slits in Young's double slit experiment is I₀Calculate the minimum separation between the points on the screen, where intensities are 2I₀ and I₀ If fringe width is b

$\frac{b}{5}$

$\frac{b}{8}$

$\frac{b}{12}$

$\frac{b}{4}$

Solution

$\frac{b}{12}$

We know that, intensity,

I = 4 l₀ cos² $\frac{ϕ}{2}$

Case (i)

2 I₀ = 4 I₀ cos² $\frac{ϕ}{2}$

$ϕ = \frac{2 π}{λ} ∆ x$

$\frac{π}{2} = \frac{2 π}{λ}$ Δx

$\frac{π}{2} = \frac{2 π}{λ} \frac{Y_{1}}{D}$ $[∆ x = \frac{Y_{1}}{D} and ϕ = \frac{π}{2}]$

⇒ $Y_{1} = \frac{λ D}{4 d}$ .....(i)

Case (ii)

I₀ = 4I₀ $\cos^{2} \frac{ϕ}{2}$

∴ $ϕ = \frac{2 π}{λ}$ Δx

$\frac{2 π}{3} = \frac{2 π}{λ} \frac{Y_{2} d}{D}$ $[∵ ϕ = \frac{2 π}{3} and ∆ x = \frac{y_{2} d}{D}]$

Y₂ = $\frac{λ D}{3 d}$

∴ ΔY = Y₂ $-$ Y₁

= $\frac{λ D}{3 d} - \frac{λ D}{4 d}$

= $\frac{λ D}{12 d}$

$∵ \frac{λD}{d} = b$

ΔY = $\frac{b}{12}$

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