8 points)Antireflection coating can be used on the eyeglasses to reduce the reflection of light: a) A 100nm thick coating is applied to the lens. What must be the coating’s index of refraction to be most effective at 500nm? (Assume the coating index of refraction is less than that of the lens). b) If the index of refraction of the coating is 1.20, find the necessary thickness of the coating at 500nm.

- the coating’s index of refraction is 1.25

- the required thickness is 104.1667 nm

Explanation:

Given the data in the question;

Thickness of coating t = 100 nm

wavelength λ = 500nm

we know that refractive index is;

t = λ/4n

make n, the subject of formula

t4n = λ

n = λ / 4t

we substitute

n = 500 / ( 4 × 100 )

n = 500 / 400

n = 1.25

Therefore, the coating’s index of refraction is 1.25

2)

given that;

Index of refraction of the coating; n = 1.20

λ = 500 nm

thickness of coating t = ?

t = λ / 4n

we substitute

t = 500 / ( 4 × 1.2 )

t = 500 / 4.8

t = 104.1667 nm

Therefore, the required thickness is 104.1667 nm

a

explanation:

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