Q. 29

# In a Young’s double slit experiment, the separation between the slits = 2.0 mm, the wavelength of the light = 600 am and the distance of the screen from the slits = 2.0 m. If the intensity at the center of the central maximum is 0.20 W m^{-2}, what will be the intensity at a point 0.5 cm away from this center along the width of the fringes?

Answer :

Given, separation between the slits

The distance between the screen and the slit,

Wavelength of light,

Let the intensity of each slit be and the corresponding amplitude be .

At the central maximum, both waves are in constructive interference and hence the intensity() is maximum and the amplitude is .

Therefore,

At point , the path difference is given by

Therefore, the phase difference is

Therefore, the resultant amplitude becomes,

Hence, the amplitude is same as a.

Therefore, the intensity would be

.

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HC Verma - Concepts of Physics Part 1

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