A thin film of soap with n = 1.37 hanging in the air reflects dominantly red light with λ = 696 nm. What is the minimum thickness of the film?

Respuesta :

Answer:

The thickness is [tex]t = 1.273 *10^{-7} \ m[/tex]  

Explanation:

From the question we are told that

     The  refractive index of the film  is  [tex]n = 1.37[/tex]

      The wavelength is  [tex]\lambda = 696 \ nm = 696 *10^{-9 } \ m[/tex]

Generally the condition for constructive interference in a film is mathematically represented as

        [tex]2 * t = [m + \frac{1}{2} ] \lambda_k[/tex]

Here t is the thickness of the film , m is the order number (0, 1, 2, 3 ... )

[tex]\lambda _k[/tex] is the wavelength of light that is inside the film , this is mathematically evaluated as

       [tex]\lambda _k = \frac{ \lambda }{ n}[/tex]

       [tex]\lambda _k = \frac{ 696 *10^{-9}}{ 1.37}[/tex]

      [tex]\lambda _k = 5.095 *10^{-7 } \ m[/tex]

So  for  m =  0

     [tex]t = [ 0 + \frac{1}{2} ] \lambda _k * \frac{1}{2}[/tex]

substituting values  

  [tex]t = [ 0 + \frac{1}{2} ] (5.095 *10^{-7}) * \frac{1}{2}[/tex]  

  [tex]t = 1.273 *10^{-7} \ m[/tex]