AP Physics 2 Quiz: Thin Film Interference

What this quiz covers

This quiz focuses on Thin Film Interference, giving you a quick way to practice the rules, question types, and explanations that matter most for AP Physics 2.

How to use this quiz

Try each quiz question before looking at the correct answer. Use the explanations to review missed ideas, then come back to similar questions until the pattern feels familiar.

Question 1

1. $t=\dfrac{m\lambda}{2n_{\text{film}}}$ (correct answer)
2. $t=\dfrac{(m+\tfrac12)\lambda}{2n_{\text{film}}}$
3. $t=\dfrac{\lambda}{n_{\text{film}}}$ only
4. Independent of $t$ because the substrate index matches air

Explanation: This problem involves thin-film interference. Light reflecting from air-film (lower to higher index) has a π phase shift, while reflection from film-substrate with matching indices has no phase shift (no index change means no reflection, but if considering a slight mismatch, higher to lower gives no phase shift). With different phase shifts at the two interfaces, there's already a π phase difference. For destructive interference (minimum reflection), the path difference must not add additional phase: 2n_film·t = mλ, giving t = mλ/(2n_film). Choice B incorrectly adds a half-wavelength, not recognizing that the existing π phase difference from reflections already sets up destructive interference. When one interface has a phase shift and the other doesn't, minimize reflection with 2nt = mλ.

Question 2

1. $2n_{\text{film}}t=(m+\tfrac12)\lambda$ (correct answer)
2. $2n_{\text{film}}t=m\lambda$
3. $t=\dfrac{\lambda}{2}$ in air, independent of $n_{\text{film}}$
4. Constructive occurs when the reflected intensity is zero

Explanation: This problem involves thin-film interference. Light reflecting from air-film (lower to higher index) has a $π$ phase shift, while reflection from film-air (higher to lower index) has no phase shift. The path difference is $2nt$, and there's already a $π$ phase difference from the different reflection conditions. For constructive interference, the path difference must add another $π$ phase to cancel the reflection phase difference: $2n_{\text{film}} t = (m + \tfrac{1}{2}) \lambda$. Choice B incorrectly uses $2n_{\text{film}} t = m \lambda$, which would result in destructive interference when the two reflections have different phase shifts. When one interface has a phase shift and the other doesn't, constructive interference requires the half-wavelength condition: $2nt = (m + \tfrac{1}{2}) \lambda$.

Question 3

1. Independent of $t$ because the film index is smallest
2. $t=\dfrac{\lambda}{n_{\text{film}}}$ only
3. $t=\dfrac{m\lambda}{2n_{\text{film}}}$ (correct answer)
4. $t=\dfrac{(m+\tfrac{1}{2})\lambda}{2n_{\text{film}}}$

Explanation: This problem involves thin-film interference. When the film has a lower index than both surrounding media, reflection at the top surface (higher to lower index) has no phase shift, while reflection at the bottom surface (lower to higher index) has a $π$ phase shift. The path difference is $2nt$, and there's already a $π$ phase difference from the reflections. For destructive interference, we need the path contribution to be in phase with this existing $π$ difference: $2n_{\text{film}} t = m \lambda$, giving $t = \dfrac{m \lambda}{2 n_{\text{film}}}$. Choice B incorrectly adds an extra half-wavelength, not recognizing that the $π$ phase difference from reflections already exists. When one reflection has a phase shift and the other doesn't, use $2nt = m \lambda$ for destructive interference.

Question 4

1. $t=\dfrac{\lambda}{4n_{\text{film}}}$ (correct answer)
2. $t=\dfrac{3\lambda}{4n_{\text{film}}}$
3. Any $t$, because both reflections undergo the same phase change
4. $t=\dfrac{\lambda}{2n_{\text{film}}}$

Explanation: This problem involves thin-film interference. Light reflecting from the air-film interface (lower to higher index) undergoes a π phase shift, while reflection from film-air (higher to lower index) has no phase shift. The path difference is 2nt, and the phase shifts differ by π, creating an effective π phase difference before considering path length. For destructive interference (minimum reflection), we need the path difference to add another π phase: 2n_film·t = (m+½)λ, giving t = (2m+1)λ/(4n_film), with minimum at m=0: t = λ/(4n_film). Choice A incorrectly suggests t = λ/(2n_film), missing that different phase shifts at the two interfaces require the quarter-wavelength condition for the first minimum. Remember: when phase shifts differ by π, destructive interference needs 2nt = (m+½)λ.

Question 5

1. Constructive interference occurs for all thicknesses because one reflection flips phase
2. Constructive interference depends only on incident intensity, not on $t$
3. Constructive interference requires $2n_{\text{film}}(2t)=m\lambda_0$ because phase shifts can be ignored
4. Constructive interference occurs when $2n_{\text{film}}(2t)=(m+\tfrac{1}{2})\lambda_0$ (correct answer)

Explanation: This problem involves thin-film interference. With the film index between air and glass, only the top reflection has a π phase shift. The path difference for thickness 2t is 2n_film(2t) = 4n_film·t. For constructive interference with one phase shift, we need the path difference to equal (m+½)λ₀, giving 2n_film(2t) = (m+½)λ₀. Choice D incorrectly ignores phase shifts entirely, while choices A and C misunderstand how thickness affects interference. When film thickness doubles, the path difference doubles, but the phase shift conditions remain unchanged. Apply the same phase analysis regardless of thickness value.

Question 6

1. $t=\lambda$ (in air), independent of $n_{\text{coat}}$
2. $2n_{\text{coat}}t=(m+\tfrac12)\lambda$ (correct answer)
3. Constructive interference occurs when the reflected intensity is smallest
4. $2n_{\text{coat}}t=m\lambda$

Explanation: This problem involves thin-film interference. Light reflecting from air-coating (lower to higher index) has a π phase shift, while reflection from coating-plastic (higher to lower index) has no phase shift. The path difference is 2nt, and there's a π phase difference from the different reflection conditions. For constructive interference, the path difference must add another π phase to make the total phase difference 2π (or 0): 2n_coat·t = (m+½)λ. Choice B incorrectly suggests 2n_coat·t = mλ, which would give destructive interference when one reflection has a phase shift and the other doesn't. When reflections have different phase shifts, constructive interference needs 2nt = (m+½)λ.

Question 7

1. No path condition; only the film's reflectivity determines minima
2. $2n_{\text{film}}t=m\lambda$
3. $n_{\text{film}}t=\lambda$ only
4. $2n_{\text{film}}t=(m+\tfrac{1}{2})\lambda$ (correct answer)

Explanation: This problem involves thin-film interference. With the film index between air and glass, reflection from air-film (lower to higher) has a $π$ phase shift, while film-glass (lower to higher) also has a $π$ phase shift. Both reflections having the same phase shift means they effectively cancel, leaving only the path difference. For minimum reflection (destructive interference), we need the path difference to create a $π$ phase shift: $2n_{\text{film}} \cdot t = (m + \tfrac{1}{2}) \lambda$. Choice B incorrectly suggests $2n_{\text{film}} \cdot t = m \lambda$, which would give constructive interference when both reflections have identical phase shifts. When both interfaces produce the same phase shift, destructive interference requires the half-wavelength condition: $2nt = (m + \tfrac{1}{2}) \lambda$.

Question 8

1. $2n_{\text{film}}t=(m+\tfrac{1}{2})\lambda_0$ (correct answer)
2. Constructive interference occurs only if the two reflected rays have equal intensities
3. Constructive interference occurs when the film is thicker than the wavelength in air, regardless of $t$
4. $2n_{\text{film}}t=m\lambda_0$

Explanation: This problem involves thin-film interference. The top reflection (air to film) produces a π phase shift since n_film > n_air, while the bottom reflection (film to substrate) has no phase shift since n_film > n_substrate. With one π phase shift, the reflected rays start π out of phase. The path difference is 2n_film·t. For constructive interference with rays initially π out of phase, we need an additional π phase from the path difference: 2n_film·t = (m+½)λ₀. Choice A incorrectly assumes no net phase shift or misapplies the interference condition. Track phase shifts systematically: low-to-high gives π shift, high-to-low gives no shift.

Question 9

1. $2n_{\text{film}}t=(m+\tfrac{1}{2})\lambda_0$ (correct answer)
2. $2n_{\text{film}}t=m\lambda_0$
3. Minimization occurs when the incident intensity is halved, independent of $t$
4. Minimization occurs for all $t$ because both reflections flip phase equally

Explanation: This problem involves thin-film interference. Both reflections produce π phase shifts: the top (air to film) because n_film > n_air, and the bottom (film to liquid) because n_liquid > n_film. With two π phase shifts, the net relative phase shift is zero (2π = 0 mod 2π). The path difference is 2n_film·t. For destructive interference when rays start in phase, we need the path difference to create a half-cycle difference: 2n_film·t = (m+½)λ₀. Choice B incorrectly applies the integer wavelength condition, which would give constructive interference. When both reflections have phase shifts, they cancel out in determining relative phase.

Question 10

1. Constructive interference occurs when the coating increases reflected intensity, not from thickness
2. $2n_{\text{coat}}t=m\lambda_0$ (correct answer)
3. $2n_{\text{coat}}t=(m+\tfrac{1}{2})\lambda_0$
4. Constructive interference occurs only if $n_{\text{coat}}=n_{\text{glass}}$, independent of $t$

Explanation: This problem involves thin-film interference. Both reflections occur at boundaries where light goes from lower to higher refractive index (air to coating and coating to glass), so both reflected rays undergo π phase shifts. Since both rays have the same phase shift, their relative phase is unchanged by reflection. The path difference is 2n_coat·t. For constructive interference when both rays start in phase, we need the path difference to equal an integer number of wavelengths: 2n_coat·t = mλ₀. Choice A incorrectly assumes a net phase shift exists, failing to recognize that two identical phase shifts cancel out. Remember to count all phase shifts and determine the net relative phase change.

Question 11

1. $2n_{\text{film}}t=m\lambda_0$
2. $2n_{\text{film}}t=(m+\tfrac{1}{2})\lambda_0$ (correct answer)
3. Maximization occurs when the film absorbs less light, independent of $t$
4. Maximization occurs only if the substrate index equals the film index, regardless of $t$

Explanation: This problem involves thin-film interference. The top reflection (air to film) has a π phase shift since n_film > n_air, while the bottom reflection (film to substrate) has no phase shift since n_film > n_substrate. With one π phase shift, reflected rays start π out of phase. The path difference is 2n_film·t. For maximum reflected light (constructive interference) with initial π phase difference, we need an additional π from path difference: 2n_film·t = (m+½)λ₀. Choice A incorrectly uses the integer condition, which would minimize reflection. When one boundary produces a phase shift, half-integer wavelengths give constructive interference.

Question 12

1. $t=\dfrac{\lambda_0}{4n_{\text{coat}}}$ (correct answer)
2. $t=0$ because destructive interference requires no path difference
3. $t=\dfrac{\lambda_0}{2n_{\text{coat}}}$
4. $t=\dfrac{\lambda_0}{4}$ because the film index does not matter

Explanation: This problem involves thin-film interference. With the coating index between air and glass, only one reflection (either top or bottom) has a π phase shift. For the first minimum (m=0) in reflected light with one phase shift, we need 2n_coat·t = (0+½)λ₀ = λ₀/2. Solving for t gives t = λ₀/(4n_coat). Choice B incorrectly doubles the thickness, likely confusing the round-trip path with the film thickness. Choice C ignores the film's refractive index, forgetting that wavelength in the medium is λ₀/n. For anti-reflection coatings, use quarter-wave thickness accounting for the medium's refractive index.

Question 13

1. $t=\dfrac{\lambda}{2n_{\text{oil}}}$ (correct answer)
2. $t=\dfrac{\lambda}{n_{\text{oil}}}$
3. Any $t$, because only the reflected intensity matters
4. $t=\dfrac{\lambda}{4n_{\text{oil}}}$

Explanation: This problem involves thin-film interference. When light reflects from the air-oil interface (lower to higher index), there's a π phase shift, but reflection from oil-water (lower to higher index) also has a π phase shift. The path difference is 2nt (twice the film thickness in the medium), and with both reflections having the same phase shift, they effectively cancel out. For constructive interference, we need 2n_oil·t = mλ, giving t = mλ/(2n_oil), with the smallest thickness at m=1: t = λ/(2n_oil). Choice A incorrectly uses λ/(4n_oil), suggesting confusion about when quarter-wavelength conditions apply versus half-wavelength conditions. When both interfaces produce the same phase shift, use the condition 2nt = mλ for constructive interference.

Question 14

1. Any $t$, because glass reflects more strongly than the film
2. $t=\dfrac{\lambda}{2n_{\text{film}}}$
3. $t=\dfrac{m\lambda}{2n_{\text{film}}}$
4. $t=\dfrac{(m+\tfrac12)\lambda}{2n_{\text{film}}}$ (correct answer)

Explanation: This problem involves thin-film interference. Light reflecting from air-film (lower to higher index) has a π phase shift, while reflection from film-glass (lower to higher index) also has a π phase shift. With both reflections having the same phase shift, they effectively cancel, leaving only the path difference 2nt to consider. For destructive interference, we need the path difference to create a π phase shift: 2n_film·t = (m+½)λ, giving t = (2m+1)λ/(4n_film). Choice C incorrectly uses t = mλ/(2n_film), which would give constructive interference when both reflections have the same phase shift. When both interfaces produce identical phase shifts, destructive interference requires 2nt = (m+½)λ.