A measuring microscope is used to examine the interference pattern. It is found that the average distance between the centers of adjacent dark fringes is 0.480 mm. Analysis Taking into consideration the phase changes that take place upon reflection, which of the following is the condition for destructive interference of the reflected light? 2t = m + 1 2 λair, m = 0, 1, 2, ... 2t = mλair, m = 0, 1, 2, ... Solve for the thickness d of the hair. µm

 2n t = m λ₀ ,    R = 0.240 mm

Explanation:

The interference by regency in thin films uses two rays mainly the one reflected on the surface and the one reflected on the inside of the film.

The ray that is reflected in the upper part of the film has a phase change of 180º since the ray stops from a medium with a low refractive index to one with a higher regrading index,

-This phase change is the introduction of a λ/2 change

-The ray passing through the film has a change in wavelength due to the refractive index of the medium

          λ₀ = λ / n

Therefore Taking into account this fact the destructive interference expression introduces an integer phase change, then the extra distance 2t is

        2 t = (m’+ ½ + ½) λ₀ / n

        2t = (m’+1) λ₀ / n

         m = m’+ 1

        2n t = m λ₀

        With   m = 0, 1, 2, ...

Where t is the thickness of the film, n the refractive index of the medium, λ the wavelength

The thickness of a hair is the thickness of the film t

           2R = t

             R = t / 2

             R = 0480/2

              R = 0.240 mm

the ima (higher than 1m) will be greater than the ima with the height of 1m.