Subject: Applied Physics 2

Topic: Interference And Diffraction

Difficulty: Medium

For constructive interference in reflected light 2μ t cos r = (2n +1) $\frac{λ}{2}$

The visible range is from 4000- 7000 AU Substitute values of n as 0,1,2,3,4……… we will get corresponding wavelengths. Those wavelengths which fall in the visible spectra will be reflected strongly.

For normal incidence cos r =1

For $ n=0,\ λ_1 = 4 μ t= 26600 AU \\[2ex] n=1,\ λ_2 = \frac{4 μ t}{3} = 8866 AU \\[2ex] n=2,\ λ_3 = \frac{4 μ t}{5} = 5320 AU \\[2ex] n=3,\ λ_4 = \frac{4 μ t}{7} = 3800 AU \\[2ex] $

So the wavelength corresponding to 5320 AU will be reflected most strongly.