Question

A thin film of water is 3100amstrang thick.lf it illuminated by white light at normal incidence.the colour of filim in reflected light will be which?

Answer 1

Answer:

Refractive index of water=4/3

Thickness of film=3100A=$310*10^{-9}m$

For constructive interference

nt=(m+1/2)λ/2

Then the wavelengths that are strongly reflected satisfy

λ=$\frac{2nt}{(m+1/2)}$

Here m=1,2,3....

λ1=$\frac{2*1.33*310*10^{-9} }{1/2}$=1649nm(Infrared region)

λ2=$\frac{2*1.33*310*10^{-9} }{3/2}$=549nm

λ3=$\frac{2*1.33*310*10^{-9} }{5/2}$=330nm(Ultraviolet region)

Clearly, 1649 nm and 330 nm wavelengths correspond to light which is invisible to the human eye. The only strong reflection occurs at 549 nm, which corresponds to the blue-violet region of the spectrum.

The reflected light is weak in the red region of the spectrum and strong in the blue-violet region. The film will, therefore, possess a pronounced blue colour.

so finally blue color is required one.

Answer 2

hence, the colour of the film in the reflected light will be green.

here we have to find the wavelengths for the construction interference occurs.

using formula, λ = 2tμ/(m - 1/2)

where t = 3100 A° , μ = 4/3

so, λ = (2)(3100)(4/3)/(m - 1/2)

= 8266.67/(m - 1/2) A°

for constructive interference Maxima occur for the following wavelengths :

16533.3 A° (m = 1) , 5511.1 A° (m = 2) and 3306.6 A° (m = 2)

we know, visible region lies between 4000 A° to 7000 A°. so, only the maximum corresponding to m = 2 lies in the visible region. light of wavelength 5511.1 A° appears green.

hence, the colour of the film in the reflected light will be green