Question

antireflective coating
a thin film of polystyrene is used as an antireflective coating for fabulite (known as the substrate). the index of refraction of the polystyrene is 1.49, and the index of refraction of the fabulite is 2.409. (figure 1)
figure
1 of 1
light rays
air
polystyrene n = 1.49
fabulite n = 2.409
part a
what is the minimum thickness of film required? assume that the wavelength of the light in air is 470 nanometers. express your answer in nanometers.
view available hint(s)
submit
provide feedback

Explanation:

Step1: Determine Phase Changes

When light reflects from a medium with a higher refractive index, it undergoes a phase shift of $\lambda/2$. Here, air ($n=1$) to polystyrene ($n=1.49$): phase shift (since $1.49>1$). Polystyrene to fabulite ($n=2.409$): phase shift (since $2.409>1.49$). So both reflected waves (top of polystyrene and top of fabulite) have a phase shift of $\lambda/2$, which cancels out. For destructive interference (antireflective), the path difference $2t$ (where $t$ is thickness) should be an integer multiple of $\lambda_n/2$, but since phase shifts cancel, we use $2t = (m + \frac{1}{2})\lambda_n$ for constructive? Wait, no—wait, antireflective coating: we want destructive interference for reflection. Wait, let's re-examine:

Wait, the two reflected waves: first from air-polystyrene (phase shift $\lambda/2$), second from polystyrene-fabulite (phase shift $\lambda/2$). So the relative phase shift between them is $0$ (both have $\lambda/2$ shift). So the path difference $2t$ (light goes down and up through polystyrene) must cause a phase difference of $\lambda_n$ (so that total phase difference is $\lambda_n$ (from path) + 0 (from shifts) = $\lambda_n$, which for destructive interference? Wait, no: destructive interference when total phase difference is $\lambda/2, 3\lambda/2$, etc. Wait, maybe I messed up. Let's recall:

For thin film interference, the condition for destructive interference (minimizing reflection) when both reflections have phase shifts (or both don't) is $2nt = m\lambda$ (where $n$ is film refractive index, $t$ thickness, $\lambda$ wavelength in vacuum/air). Wait, no: phase shift: when $n_1 < n_2$, reflection has $\lambda/2$ shift. So air ($n=1$) < polystyrene ($n=1.49$): shift. Polystyrene ($n=1.49$) < fabulite ($n=2.409$): shift. So both reflected waves have a $\lambda/2$ shift. So the relative phase shift between them is $0$ (both shifted by $\lambda/2$). So the path difference $2nt$ (since light travels $t$ down and $t$ up, so $2t$ in film, wavelength in film is $\lambda_n = \lambda/n$) must result in a phase difference of $\lambda_n$ (so that total phase difference is $\lambda_n$ (from path) + 0 (from shifts) = $\lambda_n$, which is equivalent to $2\pi$ radians, so destructive? Wait, no—wait, destructive interference occurs when the total phase difference is an integer multiple of $\lambda$ (for waves with same phase shift). Wait, maybe the formula is: when both reflections have phase shifts (or both don't), the condition for destructive interference (minimum reflection) is $2nt = m\lambda$, and for constructive is $2nt = (m + 1/2)\lambda$. But wait, let's check the case when $m=0$:

Wait, we want minimum thickness, so $m=0$? Wait, no—wait, let's think again. The two reflected waves: wave 1 (air-polystyrene) has phase shift $\lambda/2$. Wave 2 (polystyrene-fabulite) has phase shift $\lambda/2$. So the phase difference between them is (path difference phase) + (shift difference). Shift difference is $0$ (both $\lambda/2$). Path difference phase: the wave that goes through polystyrene travels an extra distance of $2t$, so phase difference is $(2t / \lambda_n) \times 2\pi = (2t n / \lambda) \times 2\pi$ (since $\lambda_n = \lambda / n$). So total phase difference is $(4\pi n t / \lambda) + 0$. For destructive interference, this should be an integer multiple of $2\pi$ (so that the waves cancel). Wait, no—destructive interference is when phase difference is $\pi, 3\pi$, etc. Wait, I'm getting confused. Let's use the standard formula for antireflective coating when both reflections have phase shifts:

The condition for destructive interference (minimum reflection) is $2nt = (m + \frac{1}{2})\lambda$? No, wait, no—let's look up the formula. Wait, actually, when both reflected waves have a phase shift of $\lambda/2$, the relative phase shift is zero, so the path difference must be such that the total phase difference is $\pi$ (for destructive), but since path difference phase is $2\pi (2t)/\lambda_n = 4\pi n t / \lambda$. So setting $4\pi n t / \lambda = \pi + 2\pi m$ (for destructive). Divide both sides by $\pi$: $4n t / \lambda = 1 + 2m$. Then $t = \lambda (1 + 2m)/(4n)$. For minimum thickness, $m=0$: $t = \lambda/(4n)$.

Ah, yes! Because when both reflections have a phase shift (so relative phase shift 0), the path difference must cause a phase difference of $\pi$ (so total phase difference $\pi$) for destructive interference. Wait, no—wait, if both have phase shift, then the two waves are out of phase by $\pi$ from the shifts? No, wait, a phase shift of $\lambda/2$ is a phase change of $\pi$ radians. So wave 1: phase shift $\pi$. Wave 2: phase shift $\pi$. So the difference in phase shifts is 0. Then the path difference: wave 2 travels an extra $2t$ in the film. The wavelength in the film is $\lambda_n = \lambda / n$. So the phase difference from path is $(2t / \lambda_n) \times 2\pi = (2t n / \lambda) \times 2\pi = 4\pi n t / \lambda$. For destructive interference, the total phase difference (path + shift) should be $\pi, 3\pi$, etc. But since shift difference is 0, path phase difference should be $\pi + 2\pi m$. So $4\pi n t / \lambda = \pi (1 + 2m)$. Divide both sides by $\pi$: $4n t / \lambda = 1 + 2m$. Then $t = \lambda (1 + 2m)/(4n)$. For minimum thickness, $m=0$: $t = \lambda/(4n)$.

Yes, that makes sense. So the formula for minimum thickness when both reflections have phase shifts is $t = \lambda/(4n)$.

Step2: Plug in the values

Given: $\lambda = 470$ nm, $n = 1.49$ (refractive index of polystyrene).

So $t = 470 / (4 \times 1.49)$.

Calculate denominator: $4 \times 1.49 = 5.96$.

Then $t = 470 / 5.96 \approx 78.86$ nm. Wait, let's compute that: 470 divided by 5.96. Let's do 470 ÷ 5.96. 5.96 × 78 = 5.96×70=417.2, 5.96×8=47.68, total 417.2+47.68=464.88. 470 - 464.88=5.12. 5.12/5.96≈0.86. So total ≈78.86 nm. So approximately 79 nm, or more precisely, let's calculate:

470 ÷ 5.96 = (470 × 100) ÷ 596 = 47000 ÷ 596. Divide numerator and denominator by 4: 11750 ÷ 149 ≈ 78.86 nm.

Answer:

$\boxed{79}$ (or more precisely 78.9, but let's check the calculation again. Wait, 470 / (4*1.49) = 470 / 5.96 ≈ 78.859, which rounds to 79 nm. So the minimum thickness is approximately 79 nanometers.