Subject: Applied Physics 2

Topic: Interference And Diffraction

Difficulty: Medium

A soap film of RI 1.33 & thickness $6 \times 10^{-5}$ cm is viewed at angle of $35^0$ to normal. Find the λs in the visible spectrum that will be absent from the reflected light. Visible $\lambda$ range: 4000 – 7000$\mathring{A}$.

μ= 1.33, t = 6 x 10$^{-5}$ cm, i=35°

$ μ= \frac{sin \ i}{sin \ r} \\[2ex] sin \ r = \frac{sin \ i }{ μ} = 0.43 $

Therefore, r = 25.46

cos r = 0.902

Condition for darkness is 2μ t cos r = nλ

Substitute values of n as 1,2,3,4……… we will get corresponding wavelengths .those wavelengths which fall in the visible spectra will remain absent.

$ For \ n =1,\ λ_1 = 2 \times 1.33\times6 \times 10^{-5} \times 0.902 = 14395 AU \\[2ex] n=2 ,\ λ_2 = 2 \times 1.33\times6 \times 10^{-5} \times \frac{0.902}{2} = 7197 AU \\[2ex] n=3,\ λ_3 = 2 \times 1.33\times6 \times 10^{-5} \times \frac{0.902}{3} =4798 AU \\[2ex] n=4,\ λ_4 = 3598 AU \\[2ex] n=5,\ λ_5 = 2879AU $

So the wavelength corresponding to 4798 AU and 3598 AU will remain absent in the visible spectrum