In the Young's double slit experiment using a monochromatic light of wavelength , the path difference (in terms of an integer ) corresponding to any point having half the peak intensity is [2013]

Question

In the Young's double slit experiment using a monochromatic light of wavelength $λ$ , the path difference (in terms of an integer $n$ ) corresponding to any point having half the peak intensity is [2013]

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(2)

Intensity, $n$ where $I_{0}$ is the peak intensity.

Here, $I = \frac{I_{0}}{2},$ $∴$ $\frac{I_{0}}{2} = I_{0} \cos^{2} \frac{ϕ}{2},$ $∴$ $ϕ = \frac{π}{2} (2 n + 1)$

$∴$ $ϕ = \frac{π}{2}, \frac{3 π}{2}, \frac{5 π}{2}, \dots$

And path difference,

$Δ x = (\frac{λ}{2 π}) ϕ$

$∴$ $Δ x = \frac{λ}{4}, \frac{3 λ}{4}, \dots, \frac{(2 n + 1) λ}{4}$

Solution

Q.
In the Young's double slit experiment using a monochromatic light of wavelength $λ$ , the path difference (in terms of an integer $n$ ) corresponding to any point having half the peak intensity is [2013]

1 $(2 n + 1) \frac{λ}{2}$

2 $(2 n + 1) \frac{λ}{4}$

3 $(2 n + 1) \frac{λ}{8}$

4 $(2 n + 1) \frac{λ}{16}$

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Solution

Q. In the Young's double slit experiment using a monochromatic light of wavelength λ, the path difference (in terms of an integer n) corresponding to any point having half the peak intensity is [2013]

1 (2n+1)λ2

2 (2n+1)λ4

3 (2n+1)λ8

4 (2n+1)λ16