Question

In Young's double slit experiment, 5th dark fringe is obtained at a point. If a thin transparent film is placed In the path of one of waves, then 7th bright fringe is obtained at the same point. The thickness of the film in terms of wavelength λ and refractive index μ will be.

Answer 1

For fifth dark fringe we have y=(m+12)Dλd=(5+12)Dλd=11Dλ2d ........ (1)Now suppose t is the thickness of the flim. So the additional pathdifference will be μt. So the condition for bright fringe is now dyD+μt=mλnow m=7, and putting value of y from (1) we getdD11Dλ2d+μt=7λ⇒μt=7λ−112λ=3λ2⇒t=3λ2μThis is the thickness of the flim in terms of wavelength and refractive index

Answer 2

Given:

The 5th dark fringe=yn

The 7th bright fringe=yn

To find:

The thickness in terms of wavelength and refractive index

Solution:

As we know

yn= (2n - 1)λD/2d

=(2×5 - 1)λD/2d

=9λD/2d

Now for the 7th bright fringe

yn=7λD/d - shift

From the above equation

we get

shift =7λD/d - 9λD/2d

shift =5λD/2d

Now putting the shift formula

D(μ-1)t/d=5λD/2d

t=2.5λ/(μ-1)

So the relation be the t=2.5λ(μ-1)