Subject: Applied Physics 2

Topic: Interference And Diffraction

Difficulty: Medium

A soap film of RI 4/3 & thickness $1.5 \times 10^{-4}$ cm is illuminated by white light incident at an angle of $45^0$. The light reflected by it is examined by a spectroscope in which is found a dark band corresponding to a wavelength $( \lambda)$ of 5000 $ \mathring{A}$. Calculate the order of this band.

μ = $\frac{4}{3}$, t = 1.5 X 10$^{-4}$ cm, i = 45° n=?, λ = 5000 A°

$ 2μt \ cos \ r =nλ \\[2ex] μ = \frac{sin \ i }{ sin \ r} \\[2ex] \frac{4}{3} = \frac{sin \ 45° }{ sin \ r} \\[2ex] Cos \ r = 0.8469 \\[2ex] n = 2 \times 1.33 \times 1.5 \times 10^{-4} \times \frac{0.8469}{5 \times 10^{-5}} \\[2ex] n= 6.758 $

Therefore n= 6