Can you please explain your answer. How did you get blue lamda. Is it because red and green makes blue? How do I find the value of blue lambda?
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
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Physic please help! - bobpursley, Wednesday, November 14, 2007 at 7:30am
Wouldnt the thickness be blue lamda/4? ?
Bob may have meant to say green light, not blue. The wavelength requirement for minimum reflection is given at
http://hyperphysics.phy-astr.gsu.edu/hbase/phyopt/soapfilm.html
For normal incidence and first order destructive interference of reflected green light,
2 N d = lambda(green)
N is the index of refraction.
I meant green. Sorry.
To understand how to find the minimum nonzero thickness of the film, we need to first understand the concept of interference and how it affects the colors we see.
Interference occurs when two or more waves interact with each other. In this case, we have a mixture of red light and green light shining on a soap film. When the light waves encounter the soap film, they reflect off the top and bottom surfaces of the film. These reflected waves then interfere with each other.
Destructive interference occurs when two waves are out of phase and their amplitudes cancel each other out. This can result in certain colors being more prominent or even completely absent in the reflected light. In this case, we want the film to appear red in reflected light, which means that the green light needs to be cancelled out by destructive interference.
Now, let's address the concept of "blue lambda/4" that you mentioned. This is based on the idea that destructive interference occurs when waves have a path difference equal to an integer multiple of half the wavelength (λ/2). If the path difference is equal to an integer multiple of the full wavelength (λ), then constructive interference occurs instead.
In this case, we want destructive interference to cancel out the green light and leave only the red light. Since the green light has a wavelength of 551 nm, the path difference needs to be equal to an odd multiple of half that wavelength (551 nm/2). This is the minimum nonzero thickness of the film required.
To calculate this thickness, we'll use the formula for the path difference in a soap film:
Path difference = 2 * n * t
where n is the refractive index of the soap film (which is given as 1.35) and t is the thickness of the film.
We know that the path difference needs to be equal to an odd multiple of 551 nm/2 (half the wavelength of the green light). So we can set up an equation:
2 * n * t = (2 * k + 1) * (551 nm/2)
where k is an integer representing the number of half-wavelengths.
To find the minimum nonzero thickness, we want k to be the smallest integer possible that satisfies the equation. That means we want k = 0, as it results in the smallest nonzero thickness. Plugging in k = 0, we can solve for t:
2 * 1.35 * t = 1 * (551 nm/2)
t = (1 * (551 nm/2)) / (2 * 1.35)
Simplifying the equation, we find that t is equal to approximately 102.59 nm.
Therefore, the minimum nonzero thickness of the film required for destructive interference to cause it to look red in reflected light is approximately 102.59 nm.