A mixture of red light (vacuum = 661 nm) and green light (vacuum = 551 nm) shines perpendicularly on a soap film (n = 1.35) that has air on either side. What is the minimum nonzero thickness of the film, so that destructive interference causes it to look red in reflected light?
nm
To determine the minimum nonzero thickness of the film for destructive interference to cause it to look red in reflected light, we need to consider the phase difference between the red and green light waves after reflection.
The phase difference can be calculated using the equation:
Δφ = 2πΔd/λ
Where:
Δφ is the phase difference
Δd is the change in optical path length
λ is the wavelength of the light
In this case, we want the red light to be cancelled out, so the phase difference should be an odd multiple of π radians.
Since the film is very thin compared to the wavelengths of light, we can assume a thin film interference model, where the phase change upon reflection is π due to the change in refractive index.
Let's calculate the phase difference for the red light:
Δφ_red = 2πΔd/λ_red
For destructive interference, the phase difference should satisfy the condition:
Δφ_red = (2n + 1)π
Substituting the known values:
(2πΔd/λ_red) = (2n + 1)π
Cancelling out the π terms:
(2Δd/λ_red) = (2n + 1)
Rearranging the equation:
Δd = (λ_red/2)(2n + 1)
Substituting the values:
Δd = (661 nm / 2)(2(1.35) + 1)
Δd = 661 nm / 2 * 4.35
Calculating:
Δd ≈ 76.21 nm
Therefore, the minimum nonzero thickness of the film so that destructive interference causes it to look red in reflected light is approximately 76.21 nm.
To determine the minimum nonzero thickness of the soap film that would cause destructive interference and make it look red in reflected light, we need to consider the concept of thin film interference.
Thin film interference occurs when a light wave passes through a thin film and reflects off both surfaces. The waves can interfere constructively or destructively depending on the phase difference between them.
In this case, we have a mixture of red light (661 nm) and green light (551 nm) incident on the soap film. We want to find the minimum thickness of the film that causes destructive interference for the red light.
To calculate the minimum nonzero thickness, we can use the equation:
2nt = (m + 1/2)λ
Where:
- "n" is the refractive index of the soap film (in this case, n = 1.35)
- "t" is the thickness of the film
- "m" is an integer representing the order of the interference (in this case, we want the first-order minimum, so m = 1)
- "λ" is the wavelength of the red light (661 nm)
Rearranging the equation to solve for "t", we get:
t = [(m + 1/2)λ] / (2n)
Plugging in the values:
t = [(1 + 1/2)(661 nm)] / (2 * 1.35)
t = 496.3 nm / 2.7
t = 183.8 nm
Therefore, the minimum nonzero thickness of the soap film for destructive interference to make it look red in reflected light is approximately 183.8 nm.