In Young's double slit experiment, monochromatic light of wavelength 5000 A ∘ is used. The slits are 1 . 0 mm apart and screen is placed at 1 . 0 m away from slits. The distance from the centre o

Question

In Young's double slit experiment, monochromatic light of wavelength 5000 A ∘ is used. The slits are 1 . 0 mm apart and 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 ______ × 10 - 6 m .

TestHub · Accepted Answer

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

So internity at centre of screen is $4 I_{0}$ (as $\cos ϕ = 1$ ).

Intensity at distance $y$ from centre,

$I = I_{0} + I_{0} + 2 \sqrt{I_{0} I_{0}} \cos ϕ$

As we know, $I_{\max} = 4 I_{0}$ . Therefore,

$\Rightarrow \frac{I_{\max}}{2} = 2 I_{0} = 2 I_{0} + 2 I_{0} \cos ϕ$

$\Rightarrow \cos ϕ = 0$

$\Rightarrow ϕ = \frac{π}{2}$

Hence, $k Δ x = \frac{π}{2}$

$\Rightarrow \frac{2 π}{λ} (dsin θ) = \frac{π}{2}$

$\Rightarrow \frac{2}{λ} d \times \frac{y}{D} = \frac{1}{2}$

$\Rightarrow y = \frac{λ D}{4 d} = \frac{5 \times 10^{- 7} \times 1}{4 \times 10^{- 3}}$

$= 125 \times 10^{- 6}$

$= 125$

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