What is the thinnest film of a coating with n=1.42 on glass n=1.52 for which destructive interference of the red component (650 nm) of an incident white light beam in air can take place by reflection?
I am confused on the n's and I have no idea what equation to use, it seems like I don't have enough info! HELP please!
Note that the angles in the sketch are exaggerated for clarity. Interference is governed by phase differences caused by reflection and by differing path lengths.
First consider the phase difference caused by reflection. Ray 1 is reflected at an air/film interface. Since nair < nfilm, the phase of Ray 1 is flipped relative to the incident ray. Ray 2 is reflected at an film/glass interface. Since nfilm < nglass, the phase of Ray 2 is also flipped relative to the incident ray. Since both rays have flipped, Δreflection/2π = 0.
Next consider the phase difference caused by the fact that Ray 2 travels an extra distance 2t (remember that we assume that the angles involved are small). This corresponds to a phase difference (number of wavelengths) of Δtravelled/2π = 2t/λfilm, where λfilm is the wavelength of the light in the thin film. The wavelength in the film is related to the wavelength in air by nfilmλfilm = nairλair.
Since Δreflection/2π = 0, we get destructive interference when
Δtravelled/2π = ½n, where n = 1, 3, 5, … .
Substituting in our expression for Δtravelled we have
2t / (nairλair/nfilm) = ½n.
Solving for t yields,
t = ¼n(nairλair/nfilm) , where n = 1, 3, 5, … .
The thinnest film for this to occur is when n = 1, so
t = ¼(1)(1.00)(650nm)/(1.42) = 114.4nm
To solve this question, you can use the concept of thin-film interference.
In this scenario, we have an incident white light beam in air (n = 1). The light goes through a coating with refractive index n = 1.42 and reflects off a glass surface with refractive index n = 1.52.
To achieve destructive interference of the red component (650 nm), we need the path difference between the two reflections to be equal to half the wavelength.
The path difference is given by the equation:
Δx = 2t(n2 - n1)
Where:
Δx is the path difference
t is the thickness of the film
n2 is the refractive index of the glass (1.52)
n1 is the refractive index of the coating (1.42)
For destructive interference, the path difference Δx should be equal to λ/2:
Δx = (650 nm) / 2 = 325 nm
Now we can rearrange the equation to solve for t:
t = Δx / 2(n2 - n1)
Plugging in the values:
t = (325 nm) / (2(1.52 - 1.42))
Simplifying:
t = (325 nm) / (0.2)
t = 1625 nm
So, the thinnest film of the coating with n = 1.42 on glass n = 1.52 for which destructive interference of the red component (650 nm) can take place by reflection is approximately 1625 nm.
To determine the thinnest film of a coating on glass where destructive interference of the red component of light can occur, we can use the concept of thin film interference.
Thin film interference occurs when a light wave is reflected from the top and bottom surfaces of a thin film, causing interference effects due to the difference in path length traveled by the waves. For destructive interference to occur, the path difference between the waves must be an odd multiple of half the wavelength.
In this case, we are interested in the red component of light with a wavelength of 650 nm. The refractive index of the coating is given as n = 1.42, and the refractive index of the glass is n = 1.52. These refractive indices indicate how much the speed of light is reduced when passing through the respective materials.
The key equation to use is the formula for the path difference, which can be derived from the relationship between wavelengths and refractive indices:
Path difference (δ) = 2nt
where:
δ = path difference
n = refractive index of the coating
t = thickness of the film
For destructive interference to occur, the path difference (δ) should be an odd multiple of half the wavelength (λ/2). Therefore, we can write:
δ = (2nt) = (2m + 1)(λ/2)
Rearranging the equation to solve for thickness (t):
t = ((2m + 1)(λ/2)) / (2n)
where:
m = integer, representing the order of interference
We need to find the thinnest film, which corresponds to the smallest possible thickness (t). For this, we need to consider the case where m = 0 (first-order destructive interference).
Substituting the given values into the equation:
t = ((2 * 0 + 1)(650 nm / 2)) / (2 * 1.42)
Simplifying the equation:
t = (1)(650 nm / 2) / 2.84
t ≈ 114.44 nm
Therefore, the thinnest film of the coating with a refractive index of 1.42 on glass with a refractive index of 1.52, for which destructive interference of the red component of light (650 nm) can occur by reflection, is approximately 114.44 nm thick.
Note: This calculation assumes the light is incident normally (perpendicularly) on the film and neglects any other potential sources of interference.