The intensity at the maximum in a Young’s double slit experiment is . Distance between two slits is , where is the wavelength of light used in the experiment. What will be the intensity in front of one of the slits on the screen placed at a distance ?

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Solution

Q.
The intensity at the maximum in a Young’s double slit experiment is $I_{0}$ . Distance between two slits is $d = 5 λ$ , where $λ$ is the wavelength of light used in the experiment. What will be the intensity in front of one of the slits on the screen placed at a distance $D = 10 d$ [2016]

1 $\frac{3}{4} I_{0}$

2 $\frac{I_{0}}{2}$

3 $I_{0}$

4 $\frac{I_{0}}{4}$

Ans.
(2)

Here, $d = 5 λ, D = 10 d, y = \frac{d}{2}$

Resultant Intensity at $y = \frac{d}{2},$ $I_{y} = ?$

The path difference between two waves at $y = \frac{d}{2}$

$Δ x = d \tan θ = d \times \frac{y}{D} = \frac{d \times \frac{d}{2}}{10 d} = \frac{d}{20} = \frac{5 λ}{20} = \frac{λ}{4}$

Corresponding phase difference, $ϕ = \frac{2 π}{λ} Δ x = \frac{π}{2}$

Now, maximum intensity in Young’s double slit experiment,

$I_{max} = I_{1} + I_{2} + 2 I_{1} I_{2} or I_{0} = 4 I (∵ I_{1} = I_{2} = I)$

$∴$ $I = \frac{I_{0}}{4}$

Required intensity,

$I_{y} = I_{1} + I_{2} + 2 I_{1} I_{2} \cos \frac{π}{2} = 2 I = \frac{I_{0}}{2}$

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