Reflection by thin layers In the figure light is incident pe

Reflection by thin layers. In the figure, light is incident perpendicularly on a thin layer of material 2 that lies between (thicker) materials 1 and 3. (The rays are tilted only for clarity.) The waves of rays r_1 and r_2 interfere, and here we consider the type of interference to be either maximum (max) or minimum (min). The table below provides the indexes of refraction n_1, n_2, and n_3, the type of interference, and the wavelength lambda in nanometers of the light as measured in air. Give the third least thickness L.

Solution

use the below condition to find the thickness.

        2nt = (m + .5)(lambda)

here, m = 0, 1, 2, ...

For the third least thickness, m = 2,

thus, the above equation becomes as follows:

2(1.64)(t) = (2.5)(630 nm)

              t = 480.18 nm = 480 nm