Question

A jeweler wishes to show off the intrinsic brilliance of their prize diamond by giving it an antireflective coating for light of wavelength 563 nm in air. What is the minimum thickness the jeweler needs for a coating material with an index of refraction of 1.49? The index of refraction of diamond is 2.41.

Answer 1

The minimum thickness of the coating material for the diamond is approximately 317.61 nm.

Step-by-step explanation:

To calculate the minimum thickness for the antireflective coating, we can use the concept of thin film interference. The condition for destructive interference is when the path length difference between the reflected waves is half a wavelength. In this case, the light is traveling from diamond (with an index of refraction of 2.41) to the coating material (with an index of refraction of 1.49) to air.

We can use the formula:

d = λ / (2 * (n2 - n1))

Where d is the thickness of the coating, λ is the wavelength in the coating material (given as 563 nm), n1 is the index of refraction of diamond, and n2 is the index of refraction of the coating material. Plugging in the values, we have:

d = 563 nm / (2 * (1.49 - 2.41))

Calculating this, we find that the minimum thickness of the coating material is approximately 317.61 nm.