Answer to Question #181033 in Optics for Pratham

Question

A soap film of refractive index 1.43is illuminated by white light incident at an angle 300. The refracted light is examined by a spectroscope in which dark band corresponding to wavelength 6х10-7m is observed. Calculate the thickness of the film.

Answer 1

$μ = 1.43 \\ λ = 6 \times 10^{-7} \;m \\ i = 30º$

Since the dark band was found n=1 condition for darkness in reflected light is given by

$2μ \times t \times cosγ = nλ \\ μ = \frac{sini}{cosγ} \\ sinγ = \frac{sini}{μ} \\ = \frac{sin30}{1.43} \\ = 0.349 \\ sin^2γ = 0.1222 \\ 1 -cos^2γ = 0.1222 \\ cos^2γ = 0.8788 \\ cosγ = 0.93744$

Thickness is found by the equation:

$t = \frac{nλ}{2μcosγ} \\ = \frac{1 \times 6 \times 10^{-7}}{2 \times 1.43 \times 0.93744} \\ = 2.23 \times 10^{-7} \;m$