A thin film of cryolite ( nc = 1.34 ) is applied to a camera lens ( ng = 1.58 ). The coating is designed to reflect wavelengths at the blue end of the spectrum and transmit wavelengths in the near infrared. What minimum thickness gives high reflectivity at 531 nm? Answer in units of nm.

Respuesta :

Answer:

=99.07nm

Explanation:

minimum thickness

2nd = (m  - 1/2)λ

d = (m - 1/2)(λ/2n)

refractive index of the thin film, n = 1.34

minimum thickness m = 1

light wavelength λ = 531nm

d = (1 - 1/2) (531 / (2)(1.34)

d = 531/5.36

  =  99.07nm