In Young's double slit experiment, 'I' is the minimum intensity and ' I 1 ' is the intensity at a point where the path difference is 4 λ where ' λ ' is the wavel

Question

In Young's double slit experiment, 'I' is the minimum intensity and ' I 1 ​ ' is the intensity at a point where the path difference is 4 λ ​ where ' λ ' is the wavelength of light used. The ratio I / I 1 ​ is (Intensities of the two interfering waves are same) ( cos 0 ∘ = 1 , cos 9 0 ∘ = 0 )

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Let the intensity of each interfering wave be $I_{0}$ .

The phase difference $ϕ$ for a path difference $Δ x$ is given by $ϕ = \frac{2 π}{λ} Δ x$ .

For $Δ x = \frac{λ}{4}$ , the phase difference is $ϕ = \frac{2 π}{λ} \times \frac{λ}{4} = \frac{π}{2} = 9 0^{\circ}$ .

The intensity at this point is $I_{1} = I_{0} + I_{0} + 2 I_{0} cos 9 0^{\circ} = 2 I_{0} + 0 = 2 I_{0}$ .

The minimum intensity $I$ in the interference pattern occurs when the phase difference is $18 0^{\circ}$ (destructive interference).

$I = I_{0} + I_{0} + 2 I_{0} cos 18 0^{\circ} = 2 I_{0} - 2 I_{0} = 0$ .

Therefore, the ratio $\frac{I}{I _{1}} = \frac{0}{2 I _{0}} = 0$ .

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