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Standard XII

Physics

Single Slit Diffraction

The wavelengt...

Question

The wavelength of the light used in Young’s double slit experiment is I. The intensity at a point on the screen is I, where the path difference is $\frac{λ}{6}$ . If $I_{0}$ denotes the maximum intensity, then the ratio of I and $I_{0}$ is

0.866

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0.5

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0.707

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0.75

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Solution

The correct option is D 0.75
$ϕ = \frac{2 π}{λ} (Δ x)$
$= 60^{\circ}$
$I = I_{0} c o s^{2} \frac{ϕ}{2}$
$\frac{I}{I_{0}} = 0.75$ .

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The wavelength of the light used in Young’s double slit experiment is I. The intensity at a point on the screen is I, where the path difference is λ6. If I0 denotes the maximum intensity, then the ratio of I and I0 is

The correct option is D 0.75ϕ=2πλ(Δx) =60∘ I=I0cos2ϕ2 II0=0.75.

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The correct option is D 0.75
$ϕ = \frac{2 π}{λ} (Δ x)$
$= 60^{\circ}$
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