Reflection by thin layers. In the figure, light is incident perpendicularly on a

Question

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 thin layer thickness L in nanometers. Give the wavelength that is in the visible range. Number 216.54 Units nm

Explanation / Answer

Here, 2L = [(m)/2] x (lambda /n2)

Here, m is an odd mumber

=> 2 * 401 =   [(m)/2] x (lambda /1.35)

=>   For m = 5   wavelength is in the range of visible light: 390 nm < visible light < 790 nm.

=>    wavelength in the visible range =   433.08 nm

Related Questions

Navigate

• Browse (All)
• Browse R
• Subjects

---

---

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.