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

We can use the formula for minimum thickness of a coating to reduce reflectivity:

t_min = λ/(4n)

where λ is the wavelength of the light in the coating, n is the index of refraction of the coating, and t_min is the minimum thickness.

First, we need to find the wavelength of the light in the coating. Since the light is incident normally, the frequency does not change, so we can use the formula:

λ = c/f

where c is the speed of light and f is the frequency.

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

Now we can use the formula for minimum thickness:

t_min = λ/(4n)

t_min = (5.22×10^-7 m)/(4×1.375) = 9.49×10^-8 m = 94.9 nm

Therefore, the answer is B) 94.9 nm.

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