Question

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 wiewing screen is now ocupied by what had been the seventh bright side fringe before covering the slit with the mica. If the wavelenght of light is 550 nm, what is the thickness of the mica? Remember that n is the index of refraction of the material given by: n=c/v, being v the speed of light through that material.

Answer 1

To develop this problem it is necessary to apply the optical concepts related to the phase difference between two or more materials.

By definition we know that the phase between two light waves that are traveling on different materials (in this case also two) is given by the equation

$\Phi = 2\pi(\frac{L}{\lambda}(n_1-n_2))$

Where

L = Thickness

n = Index of refraction of each material

$\lambda =$ Wavelength

Our values are given as

$\Phi = 7(2\pi)$

$L=t$

$n_1 = 1.58$

$n_2 = 1$

$\lambda = 550nm$

Replacing our values at the previous equation we have

$\Phi = 2\pi(\frac{L}{\lambda}(n_1-n_2))$

$7(2\pi) = 2\pi(\frac{t}{\lambda}(1.58-1))$

$t = \frac{7*550}{1.58-1}$

$t = 6637.931nm \approx 6.64\mu m$

Therefore the thickness of the mica is 6.64μm