In interference pattern using two coherent sources of light the fringe width is
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Answer : fringe width is directly proportional to wavelength .
Explanation :- as you know , fringe width is the separation between two consecutive bright or consecutive dark .
We know, for bright , yn = nλD/d ,
And for dark , yn = (2n-1)λD/d
Here, λ is wavelength of wave , d is the separation between slits and D is the separation between slit and screen.
Now, fringe width = Y₂ - Y₁ = 2λD/d - λD/d = λD/d [ when we take bright ]
also fringe width is same for dark = 3λD/2d - λD/2d = λD/d
Finally, fringe width = λD/d , hence, fringe width is directly proportional to wavelength.
Explanation :- as you know , fringe width is the separation between two consecutive bright or consecutive dark .
We know, for bright , yn = nλD/d ,
And for dark , yn = (2n-1)λD/d
Here, λ is wavelength of wave , d is the separation between slits and D is the separation between slit and screen.
Now, fringe width = Y₂ - Y₁ = 2λD/d - λD/d = λD/d [ when we take bright ]
also fringe width is same for dark = 3λD/2d - λD/2d = λD/d
Finally, fringe width = λD/d , hence, fringe width is directly proportional to wavelength.
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Answer:
proportional to wavelength
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