First slide

Young's double slit Experiment

Question

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 $D = 10 d ?$

Difficult

A

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

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

$I_{0}$
D

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

Solution

$Here, d = 5 λ, D = 10 d, y = \frac{d}{2}$
$Resultant Intensity at y = \frac{d}{2}, I_{y} = ?$
$The path difference between two waves at y = \frac{d}{2} i s$
$Δ x = d \tan θ = d \times \frac{y}{D} = \frac{d \times \frac{d}{2}}{10 d} = \frac{d}{20} = \frac{5 λ}{20} = \frac{λ}{4}$
$Corresponding phase difference, ϕ = \frac{2 π}{λ} Δ x = \frac{π}{2} .$
$Now, maximum intensity in Young's double slit experiment,$
$\begin{array}{l} I_{\max} = I_{1} + I_{2} + 2 I_{1} I_{2} \\ I_{0} = 4 I & (∵ I_{1} = I_{2} = I) \\ ∴ & I = \frac{I_{0}}{4} \end{array}$
intensity at y is $I_{y} = I_{1} + I_{2} + 2 I_{1} I_{2} \cos \frac{π}{2} = 2 I = \frac{I_{0}}{2}$

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