Monochromatic light with a wavelength of 600 km falls on the soap film in the direction of norms to its surface.

Monochromatic light with a wavelength of 600 km falls on the soap film in the direction of norms to its surface. The reflected light from the film is maximized due to interference. Determine the minimum film thickness. Refractive index of soapy water n = 1.30

The solution of the problem:
L = 600 nm = 600 * 10 ^ (- 9) nm.
n = 1.3.
1.v = 2 * d * (n ^ 2 – (sin (a)) ^ 2) ^ (1/2) + (L / 2) is the optical path difference formula, where d is the soap film thickness, a is the angle the deviation of the beam from the normal.
By the condition of the problem, a = 0, that is, v = 2 * d * n + L / 2.
Maximum condition: v = k * L.
Then: 2 * d * n + L / 2 = k * L. We express d: d = L * (k – 0.5) / (2 * n).
dmin is obtained when k = 1, that is, dmin = L * (1 – 0.5) / (2 * n) = L / (4 * n) = 600 * 10 ^ (- 9) / (4 * 1.3 ) = 1.15 * 10 ^ (- 7) m.
Answer: dmin = 1.15 * 10 ^ (- 7) m.