A coating is being applied to reduce the reflectivity of a pane of glass to light with a frequency of 5.75 × 10¹⁴ Hz that is incident normally on the pane. If the material has an index of refraction of 1.375 and the glass has an index of refraction of 1.537, what is the minimum thickness the coating should have? (c = 3.00 × 10⁸ m/s) A) 60.0 nm B) 94.9 nm C) 145 nm D) 65.2 nm E) 80.1 nm

Question

Quizplus · Accepted Answer

To reduce the reflectivity, we need to use an anti-reflective coating. The minimum thickness of the coating can be found using the formula:

t = λ/(4n)

where t is the thickness of the coating, λ is the wavelength of the incident light, and n is the index of refraction of the coating material.

First, we need to find the wavelength of the incident light in meters:

λ = c/f = 3.00 × 10^8 m/s / 5.75 × 10^14 Hz = 5.22 × 10^-7 m

Next, we can use the formula and plug in the values:

t = λ/(4n) = (5.22 × 10^-7 m) / (4 x 1.375) = 9.49 × 10^-8 m = 94.9 nm

Therefore, the minimum thickness of the coating should be 94.9 nm. The answer is option B.

A Coating Is Being Applied to Reduce the Reflectivity of a Pane