You are designing a thin transparent reflective coating for the front surface of a sheet of glass. The index of refraction of the glass is 1.52, and when it is in use the coated glass has air on both sides. Because the coating is expensive, you want to use a layer that has the minimum thickness possible, which you determine to be 104 nmnm. Part A What should the index of refraction of the coating be if it must cancel 500-nmnm light that hits the coated surface at normal incidence

1.32

Explanation:

Index of refraction of the glass = 1.52

Thickness = 104 nm

Length = 550 nm

 Using formula of index

n = L/4t

Where, L = length

t = thickness

Substituting the values into the formula we get

n = 500/(4×104)

n= 1.32

Hence, The index of refraction of the coating  is 1.32.

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