First slide

In YDSE one of the slits is covered with a thin transparent sheet

Question

The maximum intensity in Young's double slit experiment is $I_{0} .$ Distance between the slits is $d = 5 λ,$ where $λ$ is the wavelength of monochromatic light used in the experiment. What will be the intensity of light in front of one of the slits on a screen at a distance D = 10d?

Moderate

A

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

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

$I_{0}$
D

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

Solution

Suppose P is a point in front of one slit at which intensity is to be calculated from figure it.is clear that $x = \frac{d}{2} .$ Path difference between the waves reaching at P
$d = \frac{xd}{D} = \frac{(\frac{d}{2}) d}{10 d} = \frac{d}{20} = \frac{5 λ}{20} = \frac{λ}{4}$
Hence corresponding phase difference
$ϕ = \frac{2 π}{λ} \times \frac{λ}{4} = \frac{λ}{2}$
Resultant intensity at P
$I = I_{max} \cos^{2} \frac{ϕ}{2} = {Icos}^{2} (\frac{π}{4}) = \frac{I_{0}}{2}$

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