Solving for Mica Thickness in Double Slit Interference Arrangement

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Homework Help Overview

The problem involves determining the thickness of a mica flake used in a double slit interference arrangement, where the introduction of the mica changes the position of the interference fringes observed on a screen. The context includes the refractive index of mica and the wavelength of light used.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning

Approaches and Questions Raised

  • Participants discuss the need to find the path or phase difference introduced by the mica. There are attempts to relate the path difference to the position of the bright fringe and the conditions for central maxima.

Discussion Status

Some participants have provided insights into calculating the phase difference and have suggested starting with the scenario without the mica to establish a baseline. There appears to be a collaborative effort to clarify the problem setup and explore relevant equations.

Contextual Notes

Participants are working with specific values such as the refractive index of mica and the wavelength of light, but there may be assumptions regarding the setup of the double slit experiment that are not fully articulated.

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

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

Now ,given condition [tex]y= \frac{5 \lambda D}{d}[/tex]

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

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