A parallel beam of white light falls on a thin film whose refractive i...
Problem
A parallel beam of white light falls on a thin film whose refractive index is equal to 4/3. The angle of incidence i=53 degree. What must be the minimum film thickness if the reflected light is to be coloured yellow (lambda of yellow=0.6m) most intensively?(tan53degree=4/3)?
Solution
When white light is incident on a thin film, it undergoes reflection and transmission from both the surfaces of the film. The reflected and transmitted light waves interfere with each other, leading to the formation of colored fringes. The condition for constructive interference for a particular wavelength λ is given by:
2nt = (m + 1/2) λ
where n is the refractive index of the film, t is the thickness of the film, m is an integer, and λ is the wavelength of light.
We know that the wavelength of yellow light (λ) is 0.6 micrometers. We need to find the minimum thickness of the film for which the reflected yellow light is most intense.
To find the minimum thickness of the film, we need to use the condition for constructive interference for yellow light. Let the minimum thickness of the film be t. Then,
2nt = (m + 1/2) λ
2 × (4/3) × t = (m + 1/2) × 0.6 × 10^-6
8t/3 = (m + 1/2) × 0.6 × 10^-6
To get the most intense reflection of yellow light, we need to choose the smallest possible value of m, which is zero. Therefore,
8t/3 = 0.3 × 10^-6
t = 0.1125 micrometers (minimum thickness of the film)
Answer
The minimum thickness of the film required to reflect yellow light most intensively is 0.1125 micrometers.