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 $\frac{λ}{6} (λ$ being the wavelength of the light used) is I .If I_odenotes the maximum intensity $\frac{I}{I_{0}}$ is equal to

Moderate

A

$\frac{1}{2}$
B

$\frac{3}{4}$
C

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

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

Solution

The resultant intensity at a point where the path difference is x is given by I_RES= 4 I₀COS² $(\frac{π X}{λ})$
The resultant intensity is given by
$I = I_{1} + I_{2} + 2 \sqrt{I_{1} I_{2}} \cos δ I_{resultant} = I + I + 2 Icos (π / 3) (∵ δ = \frac{2 π}{λ} Δx = \frac{2 π}{λ} \times \frac{1}{6} = \frac{π}{3}) I_{resultant} = 3 I I_{max} = 4 I Therefore, \frac{I_{resultant}}{I_{max}} = \frac{3}{4}$

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