First slide

Young's double slit Experiment

Question

In a Young’s double slit experiment the intensity at a point where the path difference is $λ$ /6 ( $λ$ being the wavelength of light used) is i. If $I_{0}$ denotes the maximum intensity, $I / I_{0}$ is equal to

Moderate

A

3 / 4
B

$\frac{1}{\sqrt{2}}$
C

$\frac{\sqrt{3}}{2}$
D

1 / 2

Solution

$\begin{array}{l} Δ x = λ / 6 \\ ϕ = 60^{o} \\ I = I_{0} \cos^{2} \frac{60}{2} = I_{0} C o s^{2} 30 = \frac{3 I_{6}}{4} \frac{I_{}}{I_{0}} = \frac{3}{4} \end{array}$

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