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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