Question

A soap film(n=1.33) in air is 320nm thick.if it is illuminated with white light at normal incidence,what colour will it appear to be in reflected light???

Answer 1

 λ = light ray's wavelength :  380 nm  to  760 nm  for visible light (white rays).

 μ_film = 1.33
 thickness of film = d = 320 nm
 θ1 = 0°  = angle of incidence at the top layer of film
 μ_air = 1
 θ2 = angle of refraction into the soap film = angle of reflection at the film-air interface.
    μ_air Sin θ1  = μ_film  Sin θ2          --- (1)

   There is a phase shift of 180 deg. when light rays are reflected at the top surface of the film, as μ_air < μ_film.  There is no phase shift when rays are reflected at the  film-to-air interface.  The OPD (optical path difference) is:
    
   2 μ_film  d  Cos θ2  = (m - 1/2) λ   for constructive interference.    --(2)
   2 μ_film  d  Cos θ2  = m λ    for destructive interference.    -- (3)
                   here m = an integer.

  As θ1 = 0°,  θ2 = 0°.
      2 * 1.33 * 320 nm  = Optical path difference OPD 
                 = 661.12  nm

   In equations  (2) and (3),  the RHS is λ/2, λ, 3λ/2 ...  We have the range of white light wavelengths mentioned above.
  661.12 = λ for red color  satisfies equation  (3).  So red color is destroyed.
  661.12 = 1.5 λ for  λ = 440.7 nm satisfies equation (4),   ie., violet color is constructively added.
  

Answer 2

It will appear in green color.

Step by step explanation

Given data:

refractive index of air$=1.33$

thickness of the soap film$=320nm$

To find:

We need to determine the color it will appear in reflected light.

Formula to be used:

$2$μ$d=(n+1/2)$λ

where:μ$=$$1.33$ (refractive index)

$d=$thickness

λ$=$wavelength

$n=1,2,3...$

Solution:

Since constructive interference happen we use the formula:

$2$μ$d=(n+1/2)$λ

taking n as $1$;

λ$=2$μ$d/(n+1/2)$

$=2$×$1.33$×$320$×$10^{-9}$$/(1+1/2)$

$=$$5.67$×$10^{-7}$

$=567nm$

This wavelength appears in green region of the spectrum hence the color the film will be in reflected light is green.

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