Solving for Mica Thickness in Double Slit Interference Arrangement

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The discussion centers on calculating the thickness of a mica flake used in a double slit interference arrangement, where the mica alters the interference pattern. The mica has a refractive index of 1.58 and covers one slit, shifting the central point to the previously fifth bright fringe with a light wavelength of 572 nm. Participants suggest starting by calculating the path difference or phase difference without the mica to establish a baseline for the interference pattern. The relevant equations involve the path difference and the relationship between fringe positions and wavelength. The conversation concludes with appreciation for the assistance provided in solving the problem.
Callisto
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Double slit arrangment??

Can anybody help me with this problem?

A thin flake of mica(n=1.58) is used to cover one slit of a double slit interference arrangment.The central point on the viewing screen is now covered by what had been the 5th bright fringe before the mica was used. Find the thickness of the mica if light of wavelength 572nm is used?

Where do i start with such a problem?
Any hints or tips are very much appreciated.

Callisto :biggrin:
 
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The problem is that of finding the path difference or phase difference between the two systems. Calculate the diffraction problem without the mica involved. From there you will know the phase and can go on to solve the other part.

dt
 
the path differnce will be

[n-1]t - \frac{dy}{D}

Now ,given condition y= \frac{5 \lambda D}{d}

and [n-1]t - \frac{dy}{D} =0 for central maxima/point
:smile:
 
Thankyou for your replies,
they where of great assistance
 
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