Question

a plastic film (n = 1.49) reflects light, and the brightest color you see is orange light at 606 nm. if that is from the longest (m = 1) wavelength possible, how thick is the plastic film at that point?? nm

Explanation:

Step1: Recall the formula for thin - film interference (constructive interference for reflection)

For a thin film, when light reflects from the top and bottom surfaces, the condition for constructive interference (bright fringes) in the case of a film with \(n>1\) (plastic film, \(n = 1.49\)) and considering the phase changes: when light reflects from a medium with higher refractive index, a phase change of \(\lambda/2\) occurs. For a plastic film in air (\(n_{air}=1\)), the reflection from the top surface (air - plastic, \(n\) increases) has a phase change of \(\lambda/2\), and the reflection from the bottom surface (plastic - air, \(n\) decreases) has no phase change. So the path difference for constructive interference is \(2nt=(m +\frac{1}{2})\lambda\) when \(m = 0,1,2,\cdots\). But in this problem, we are told that it is the longest wavelength possible for \(m = 1\). Wait, actually, let's re - examine the condition. Wait, maybe I made a mistake. Let's think again.

Wait, the formula for constructive interference in thin - film reflection: when the film is between two media with lower refractive index (air on both sides? No, the film is a plastic film, so air on top and maybe air on the bottom? Wait, the problem is about reflection. So the light is incident from air (\(n = 1\)) onto the plastic film (\(n=1.49\)) and then reflects. The reflection from the first surface (air - plastic) has a phase shift of \(\lambda/2\) (because \(n_{plastic}>n_{air}\)), and the reflection from the second surface (plastic - air) has no phase shift (because \(n_{air}<n_{plastic}\)). The path difference for the two reflected waves is \(2nt\) (the light travels down \(t\) and up \(t\) in the film, so total path length in the film is \(2t\), and the optical path length is \(2nt\)). The phase difference between the two reflected waves is due to the path difference and the phase shift. For constructive interference, the total phase difference should be an integer multiple of \(\lambda\).

The phase shift from the first reflection is \(\lambda/2\), and the phase shift from the path difference is \(\frac{2nt}{\lambda}\times2\pi\). The total phase difference for constructive interference is \(\frac{2nt}{\lambda}\times2\pi+\pi = 2\pi m\) (where \(m = 1,2,3,\cdots\)) (because the first reflection has a \(\pi\) phase shift (\(\lambda/2\) path equivalent) and the path difference gives a phase shift of \(\frac{2nt}{\lambda}\times2\pi\)). Dividing both sides by \(2\pi\), we get \(\frac{2nt}{\lambda}+\frac{1}{2}=m\). So \(2nt=(m - \frac{1}{2})\lambda\).

But the problem says "the longest wavelength possible" for \(m = 1\). Wait, the wavelength \(\lambda\) in the formula is the wavelength in vacuum. Let's solve for \(t\): \(t=\frac{(m-\frac{1}{2})\lambda}{2n}\).

We are given \(m = 1\), \(\lambda=606\) nm, \(n = 1.49\).

Step2: Substitute the values into the formula

Substitute \(m = 1\), \(\lambda = 606\) nm, and \(n=1.49\) into the formula \(t=\frac{(m-\frac{1}{2})\lambda}{2n}\)

First, calculate \(m-\frac{1}{2}\) when \(m = 1\): \(1-\frac{1}{2}=\frac{1}{2}\)

Then, \(2n = 2\times1.49 = 2.98\)

Now, \(t=\frac{\frac{1}{2}\times606}{2.98}\)

Calculate the numerator: \(\frac{1}{2}\times606 = 303\)

Then, \(t=\frac{303}{2.98}\approx101.68\) nm? Wait, that can't be right. Wait, maybe I mixed up the formula.

Wait, another approach: maybe the formula is \(2nt=m\lambda\) for constructive interference. Wait, let's think about the phase changes again. If the film is in air, and we have reflection from the top (phase change \(\lambda/2\)) and reflection from the bottom (no phase change). So the two reflected waves have a phase difference of \(\pi\) (from the phase changes) plus the phase difference from the path difference \(2nt\). For constructive interference, the total phase difference should be an integer multiple of \(2\pi\). So \(\pi+\frac{2nt}{\lambda}\times2\pi = 2\pi m\). Divide both sides by \(\pi\): \(1+\frac{4nt}{\lambda}=2m\), so \(\frac{4nt}{\lambda}=2m - 1\), so \(2nt=(m-\frac{1}{2})\lambda\). But if we are looking for the longest wavelength for a given \(m\), we know that \(\lambda=\frac{2nt}{m - \frac{1}{2}}\). To maximize \(\lambda\), we need to minimize \(m-\frac{1}{2}\). The smallest non - negative value of \(m\) for which this makes sense: when \(m = 1\), \(m-\frac{1}{2}=\frac{1}{2}\); when \(m = 0\), \(m-\frac{1}{2}=-\frac{1}{2}\), but wavelength can't be negative. Wait, maybe the formula is different.

Wait, maybe the problem is considering that the film is a thin film where the condition for constructive interference in reflection is \(2nt=(m+\frac{1}{2})\lambda\) for \(m = 0,1,2,\cdots\), and we want the longest wavelength for \(m = 1\). Wait, no, the wavelength is inversely proportional to \(m+\frac{1}{2}\) when \(t\) and \(n\) are constant. So to have the longest wavelength, we need the smallest value of \(m+\frac{1}{2}\). The smallest \(m\) is \(m = 0\), but the problem says \(m = 1\). Wait, the problem states "the longest (m = 1) wavelength possible". So we have \(2nt=(m+\frac{1}{2})\lambda\), and we need to find \(t\) when \(m = 1\) and \(\lambda = 606\) nm.

So \(2nt=(1+\frac{1}{2})\lambda=\frac{3}{2}\lambda\)

Then, \(t=\frac{3\lambda}{4n}\)

Substitute \(\lambda = 606\) nm and \(n = 1.49\)

\(t=\frac{3\times606}{4\times1.49}=\frac{1818}{5.96}\approx305\) nm? Wait, that still doesn't seem right.

Wait, maybe the formula is \(2nt=m\lambda\) for constructive interference. Let's check the phase changes again. If there is a phase change at both surfaces (but in this case, the bottom surface is plastic - air, so no phase change). Wait, no, the top surface is air - plastic (phase change \(\lambda/2\)), bottom surface is plastic - air (no phase change). So the net phase change from reflections is \(\lambda/2\). The path difference is \(2nt\). So for constructive interference, the path difference plus the phase change should be an integer multiple of \(\lambda\). So \(2nt+\frac{\lambda}{2}=m\lambda\), where \(m = 1,2,3,\cdots\) (because when \(m = 0\), \(2nt=-\frac{\lambda}{2}\), which is not physical). So \(2nt=(m - \frac{1}{2})\lambda\)

We are given that it is the longest wavelength possible for \(m = 1\). The wavelength \(\lambda\) is related to \(t\) by \(\lambda=\frac{2nt}{m - \frac{1}{2}}\). To maximize \(\lambda\), we need to minimize \(m - \frac{1}{2}\). For \(m = 1\), \(m-\frac{1}{2}=\frac{1}{2}\). If \(m = 0\), \(m-\frac{1}{2}=-\frac{1}{2}\), but wavelength can't be negative. So with \(m = 1\), we have \(\lambda=\frac{2nt}{\frac{1}{2}} = 4nt\)? No, wait, from \(2nt=(m - \frac{1}{2})\lambda\), we can solve for \(\lambda\): \(\lambda=\frac{2nt}{m - \frac{1}{2}}\). So when \(m = 1\), \(\lambda=\frac{2nt}{1-\frac{1}{2}}=\frac{2nt}{\frac{1}{2}} = 4nt\). But we know \(\lambda = 606\) nm, so \(4nt=606\), then \(t=\frac{606}{4n}=\frac{606}{4\times1.49}=\frac{606}{5.96}\approx101.68\) nm. But this seems too small.

Wait, maybe the formula is \(2nt=m\lambda\) for destructive interference, but no, the problem is about the brightest color, which is constructive interference.

Wait, let's check a reference. The formula for constructive interference in thin - film reflection when there is one phase change (top surface) is \(2nt=(m+\frac{1}{2})\lambda\), \(m = 0,1,2,\cdots\). So if \(m = 1\), then \(2nt=(1+\frac{1}{2})\lambda=\frac{3}{2}\lambda\), so \(t=\frac{3\lambda}{4n}\)

Substitute \(\lambda = 606\) nm and \(n = 1.49\):

\(t=\frac{3\times606}{4\times1.49}=\frac{1818}{5.96}\approx305\) nm.

Wait, now I'm confused. Let's do the calculation again.

\(n = 1.49\), \(\lambda = 606\) nm, \(m = 1\)

The correct formula for constructive interference in thin - film reflection (with one phase shift at the top surface) is:

\(2nt=(m+\frac{1}{2})\lambda\), where \(m = 0,1,2,\cdots\)

We need to find \(t\), so \(t=\frac{(m+\frac{1}{2})\lambda}{2n}\)

For \(m = 1\):

\(m+\frac{1}{2}=1 + 0.5=1.5\)

\(2n=2\times1.49 = 2.98\)

\(t=\frac{1.5\times606}{2.98}=\frac{909}{2.98}\approx305\) nm.

Yes, that makes more sense. So the thickness \(t\) is given by \(t=\frac{(m +\frac{1}{2})\lambda}{2n}\)

Answer:

\(\approx305\) (The more precise calculation: \(\frac{1.5\times606}{2.98}=\frac{909}{2.98}\approx305.03\) nm)