In a Young’s double slit experiment, the intensities at two points, for the path difference λ 4 and λ 3 ( λ being the wavelength of light used) are I 1 and I 2 respectively. If I 0 denotes the intensi

Question

In a Young’s double slit experiment, the intensities at two points, for the path difference λ 4 and λ 3 ( λ being the wavelength of light used) are I 1 and I 2 respectively. If I 0 denotes the intensity produced by each one of the individual slits, then I 1 + I 2 I 0 = _ _ _ _ _ _ .

TestHub · Accepted Answer

As we know, resultant intensity is given by,

$I = 4 I_{0} \cos^{2} (\frac{Δ ϕ}{2})$ and phase difference can be written as $∆ ϕ = \frac{2 π}{λ} ∆ x$ .

For first case:

$I_{1} = 4 I_{0} \cos^{2} (\frac{2 π}{λ} \times \frac{λ}{4}) = 4 I_{0} \cos^{2} (\frac{π}{4}) = 2 I_{0}$

For second case:

$I_{2} = 4 I_{0} \cos^{2} (\frac{2 π}{λ} \times \frac{λ}{3}) = 4 I_{0} \cos^{2} (\frac{2 π}{3}) = I_{0}$

Therefore, required value of $\frac{I_{1} + I_{2}}{I_{0}} = 3$ .

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