In Young’s double slit experiment, monochromatic light of wavelength 5000is used. The slits are 1.0 mm apart and the screen is placed at 1.0 m away from slits. The distance from the centre of the screen where intensity becomes half of the maximum intensit

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Solution

Q.
In Young’s double slit experiment, monochromatic light of wavelength 5000 $Å$ is used. The slits are 1.0 mm apart and the screen is placed at 1.0 m away from slits. The distance from the centre of the screen where intensity becomes half of the maximum intensity for the first time is ____ $\times 10^{- 6} m$ . [2024]

Ans.
(125)

Let intensity of light on screen due to each slit is $I_{0}$

So, intensity at centre of screen is $4 I_{0}$

$λ = 5000 Å, d = 1 \times 10^{- 3} m$

$D = 1 m$

$Δ x = d \sin θ$ ...(i)

Let intensity of light on screen due to each slit is $I_{0}$

So intensity at centre of screen is $4 I_{0}$

$I_{\max} = 4 I_{0}$

$I_{\max} = (\sqrt{I_{0}} + \sqrt{I_{0}}) = {(2 \sqrt{I_{0}})}^{2} = 4 I_{0}$

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

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

$\frac{Δ ϕ}{2} = \frac{π}{4} or - \frac{π}{4} \Rightarrow Δ ϕ = \frac{π}{2} or - \frac{π}{2}$

$\frac{2 π}{λ} \times Δ x = Δ ϕ$

$\frac{2 π}{λ} \times Δ x = \frac{π}{2} \Rightarrow Δ x = \frac{λ}{4}$

From (i): $Δ x \approx d \tan θ, Δ x \approx \frac{d y}{D}$

$\frac{λ}{4} = \frac{d y}{D} \Rightarrow y = \frac{λ D}{4 d}$

$y = \frac{(5000 \times 10^{- 10}) (1)}{4 \times 10^{- 3}} = \frac{500}{4} \times 10^{- 6} = 125 \times 10^{- 6}$

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