A thin layer of magnesium fluoride (n = 1.38) is used to coat a flint-glass lens (n = 1.61).
What thickness should the magnesium fluoride film have if the reflection of 707-nm light is to be suppressed? Assume that the light is incident at right angles to the film.

Answers

Answer 1

The thickness of the magnesium fluoride film should be 205.7 nm to suppress the reflection of 707-nm light.

To suppress the reflection of 707-nm light, we need to create destructive interference between the waves reflected from the top and bottom surfaces of the magnesium fluoride film.

The condition for destructive interference is:

[tex]2nt = (m + 1/2)λ[/tex]

where n is the refractive index of the magnesium fluoride film, t is the thickness of the film, m is an integer representing the order of the interference, and λ is the wavelength of the light in vacuum.

In this case, we want m = 0, so the equation simplifies to:

2nt = λ/2

We are given n1 = 1.38 and n2 = 1.61, and the wavelength of light in vacuum λ = 707 nm. We can use the formula for the reflection coefficient at an interface between two media:

[tex]r = (n1 - n2)/(n1 + n2)[/tex]

to find the phase shift upon reflection at the top surface of the film. In this case, the reflection coefficient is:

r = (1.38 - 1.61)/(1.38 + 1.61) = -0.11

The phase shift is then:

δ = 2πr = -0.69π

The phase shift upon reflection at thebof the film is zero since the light is going from a higher to a lower refractive index medium. Therefore, the total phase shift upon reflection from both surfaces is:

Δ = 2δ = -1.38π

To create destructive interference, we need to adjust the thickness of the film so that the total phase shift upon reflection is an odd multiple of π. In other words:

Δ = (2n + 1)π

where n is an integer. Solving for t, we get:

[tex]t = [(2n + 1)λ/4n] / (n2 - n1)[/tex]

Plugging in the given values, we get:

[tex]t = [(2(0) + 1)(707 nm)/(4(0))] / (1.61 - 1.38) = 205.7 nm[/tex]

Therefore, the thickness of the magnesium fluoride film should be 205.7 nm to suppress the reflection of 707-nm light.

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Answer:

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Answers

Answer:

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Answer:

The answer will be :

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