White light containing the wavelengths between 420 and 720 nm traveling in the air falls on a thin layer n1 = 1.5 and a thickness of 0.000001 m. If this layer is placed on another thin film n2 = 1.4, determine the wavelengths of the light that is not reflected in the air.

The wavelength of the light that are not reflected in the air is,

 λ5= 600 nm, λ6= 500 nm, λ7=428 nm.

Explanation:

For the above problem, we have the case as no<n1 and n2 < n1 .so we have(1<1.5;1.4<1.5)To find the wavelengths of the light that is not reflected in the air.Since the phase changes are not common on both the surfaces/we have the formula,   λ=2d n²/m          ( where m=1,2,3,4,5...)solving the problem we get,λ5= 600 nm, λ6= 500 nm, λ7=428 nmThe wavelength of the light that are not reflected in the air is,

         λ5= 600 nm, λ6= 500 nm, λ7=428 nm.

Mass and velocity are the 2 factors used to calculate kinetic energy. the equation for kinetic energy is (1/2)mv^2. hope this !