Determining Mica Thickness Using Interference Fringes

[inflames829]

Homework Statement

A thin flake of mica (n= 1.58) is used to cover one slit of a double slit interference arrangement. The central point on the viewing screen is now covered by what had been the 9th bright fringe before the mica was used. What is the thickness of mica if light of wavelength 556nm is used? answer in microns

Homework Equations

The Attempt at a Solution

n lambda = xb/L
where
λ is the wavelength of the light,
b is the separation of the slits, the distance between A and B in the diagram to the right
n is the order of maximum observed (central maximum is n=1),
x is the distance between the bands of light and the central maximum (also called fringe distance), and
L is the distance from the slits to the screen centerpoint.


This is the only equation i can find so i assume you use it. Is this correct? and if so can someone give me hints on what values to put in for L and x. please. THANKYOU

 

[Kurdt]

The path difference when one slit is covered with mica will be slightly different.

\frac{xb}{L} = (n-1)t

where t is the thickness of the mica.