Question

You have explored constructive interference from multi-layer thin films. It is also possible for interference to be destructive, a phenomenon exploited in making antireflection coatings for optical elements such as eyeglasses. In order to allow the lenses to be thinner (and thus lighter weight), eyeglass lenses can be made of a plastic that has a high index of refraction (np = 1.70). The high index causes the plastic to reflect light more effectively than does glass, so it is desirable to reduce the reflection to avoid glare and to allow more light to reach the eye. This can be done by applying a thin coating to the plastic to produce destructive interference.
a. Consider a plastic eyeglass lens with a coating of thickness d with index nc . Light with wavelength is incident perpendicular to the lens. If nc < n p , then determine an equation for d in terms of the given variables (and an integer m) in order for there to be destructive interference between the light reflected from the top of the coating and the light reflected from the coating/lens interface.
b. Repeat part a assuming that nc > n p .
c. Choose a suitable value for nc and calculate a value for d that will result in destructive interference for 500 nm light. Note that materials to use for coatings that have nc < 1.3 or nc > 2.5 are difficult to find.
d. Does the index of refraction n p of the eyeglass lens itself matter? Explain.

Answer 1

Answer:

a)   d sin θ = m λ₀ / n

b)   d sin θ = (m + ½) λ₀ / n

c)    d = 2,439 10⁻⁷ m

Explanation:

For the interference these rays of light we must take as for some aspects,

* when a beam of light passes from a medium with a lower index to one with a higher index, the reflected ray has a phase change of 18º, this is equivalent to lam / 2

* when the ray penetrates the lens the donut length changes by the refractive index

            λ = λ₀ / n

now let's write the destructive interference equation for these lightning bolts

           d sin θ = (m´ + 1/2 + 1/2) λ / n = (m` + 1) λ₀ / n

           d sin θ = m λ₀ / n

b) now nc> np

in this case there is no phase change in the reflected ray and the equation for destructive interference remains

             d sin θ = (m + ½) λ₀ / n

c) select the value of nc = 2.05 of the ZnO

we calculate the thickness of the film (d)

            d = m λ / (n sin 90)

in this type of interference the observation is normal, that is, the angle is 90º)

           d = 1 500 10-9 / (2.05 1)

           d = 2,439 10⁻⁷ m

d) the lens replacement index is very important because it depends on its relation with the film index which equation to destructively use interference