46. A thin film of oil with a thickness of 90 nm rests on top of a pool of water. When white light incident on the film is reflected, what colors are seen? The refractive index of the oil is noil = 1.45.
This is my best attempt, just wondering if I was right. Any help would be appreciated.
2d = (m + ½)(λ /noil) (m = 0, assuming the minimum thickness)
(n air < n oil > n water) → the wave is inverted on the first reflection and not on the second reflection
[d = λ/4(noil) ] → 90nm = λ/4(1.45)
λ = 522 nm
According to wavelengths of visible light: Green = 520 - 565 nm
Well, it seems like you've made a great attempt at solving the problem! However, when it comes to the colors seen in the reflected light, it's not just about the wavelength of light, but also about the interference of the waves.
When white light reflects off the thin film of oil, it undergoes interference based on the thickness of the film. This interference can cause certain wavelengths of light to reinforce each other (constructive interference) or cancel each other out (destructive interference).
In this case, with a thickness of 90 nm, you calculated the wavelength of the reflected light to be 522 nm. Now, remember that white light is composed of a range of wavelengths spanning the visible spectrum.
So, instead of just seeing a single color, you would observe a range of colors due to the interference of different wavelengths. In this case, with a calculated wavelength of 522 nm, you would likely see a mix of green and blue hues.
Keep in mind that the actual colors observed can also depend on the angles of incidence and observation, as well as other factors. But generally, you can expect to see a combination of green and blue when white light reflects off the thin film of oil with a thickness of 90 nm.
Hope this clarifies things a bit, and remember, when it comes to colors, there's always room for a bit of "refraction" humor!
Your attempt is almost correct. The equation you used, 2d = (m + ½)(λ /noil), is correct for determining the wavelengths of light reflected from a thin film. However, there is a mistake in the calculation of the wavelength.
Using the equation 2d = (m + ½)(λ /noil), where d is the thickness of the film, m is an integer representing the order of the interference, λ is the wavelength of light, and noil is the refractive index of the oil, we can rearrange the equation to solve for λ:
λ = 2dnoilm + noil/2
Plugging in the values, with d = 90 nm and noil = 1.45, we get:
λ = 2 * 90 nm * 1.45 + 1.45/2
λ = 261 nm + 0.725 nm
λ = 261.725 nm
So, the calculated wavelength is approximately 261.725 nm.
To determine the colors seen, we use the fact that different colors of light correspond to different wavelengths. Here are the approximate ranges for the colors of visible light:
- Red: 620 - 750 nm
- Orange: 590 - 620 nm
- Yellow: 570 - 590 nm
- Green: 495 - 570 nm
- Blue: 450 - 495 nm
- Violet: 380 - 450 nm
Since the calculated wavelength of 261.725 nm falls within the ultraviolet range (below 380 nm), no visible color would be seen in this case.
Well done on your attempt! Your calculations are mostly correct.
To determine the colors seen when white light is reflected off the thin film of oil, we need to consider the phenomenon of interference. Interference occurs when two or more light waves combine, either constructively or destructively. In this case, we have two reflections: one at the top surface of the oil and the other at the bottom surface.
The equation you used, 2d = (m + ½)(λ / noil), is the equation for constructive interference in thin films, where d is the thickness of the film, m is the order of the interference, λ is the wavelength of the light, and noil is the refractive index of the oil.
Since you assumed the minimum thickness (m = 0), the equation simplifies to d = λ / (4 * noil). Plugging in the values given, we have:
d = 90 nm = λ / (4 * 1.45)
Rearranging the equation to solve for λ:
λ = 90 nm * (4 * 1.45) = 522 nm
So, the wavelength of the light is 522 nm.
Now, according to the wavelengths of visible light, you are correct that green light falls within the range of 520 - 565 nm. Therefore, when white light is reflected off the thin film of oil with a thickness of 90 nm, primarily green light will be seen.
However, it's important to note that a thin film of oil can produce a range of colors due to the phenomenon of constructive and destructive interference. So, in addition to green, other colors may also be observed depending on the exact thickness of the film and the incident angle of the light.
I hope this clarifies things for you! Let me know if you have any further questions.