Thin-Film Interference
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AP Physics 2 › Thin-Film Interference
A thin film with index greater than air lies on a substrate with the same index as air. Light reflects from the top and bottom film surfaces and recombines in air. For normal incidence, which thickness minimizes reflected light at wavelength $\lambda$?
Independent of $t$ because the substrate index matches air
$t=\dfrac{(m+\tfrac12)\lambda}{2n_{\text{film}}}$
$t=\dfrac{m\lambda}{2n_{\text{film}}}$
$t=\dfrac{\lambda}{n_{\text{film}}}$ only
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.
A thin film of thickness $t$ has refractive index greater than air and is in contact with air on both sides. Light reflects from the top and bottom surfaces and recombines in air. Which thickness condition produces constructive interference in reflected light (normal incidence)?
$t=\dfrac{\lambda}{2}$ in air, independent of $n_{\text{film}}$
Constructive occurs when the reflected intensity is zero
$2n_{\text{film}}t=(m+\tfrac12)\lambda$
$2n_{\text{film}}t=m\lambda$
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: $2nt = (m + \tfrac{1}{2}) \lambda$.
A thin film with refractive index less than both surrounding media is sandwiched between two higher-index materials. Reflections from the top and bottom film surfaces recombine in the top medium. For normal incidence, which thickness gives destructive interference in reflected light?
$t=\dfrac{\lambda}{n_{\text{film}}}$ only
$t=\dfrac{(m+\tfrac{1}{2})\lambda}{2n_{\text{film}}}$
$t=\dfrac{m\lambda}{2n_{\text{film}}}$
Independent of $t$ because the film index is smallest
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.
A soap film (index greater than air) has air on both sides. Light reflects from the top and bottom surfaces and recombines in air. For normal incidence with wavelength $600,\text{nm}$ in air, which thickness minimizes reflected light?
$t=\dfrac{\lambda}{2n_{\text{film}}}$
Any $t$, because both reflections undergo the same phase change
$t=\dfrac{3\lambda}{4n_{\text{film}}}$
$t=\dfrac{\lambda}{4n_{\text{film}}}$
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.
Light of vacuum wavelength $\lambda_0$ is normally incident from air onto a thin film on glass. The film’s refractive index is between that of air and glass, so only the reflection at the top surface has a $ rac{1}{2}\lambda$ phase shift. Two reflected rays recombine in air. If the film thickness is doubled from $t$ to $2t$, which statement best describes when constructive interference occurs?
Constructive interference occurs for all thicknesses because one reflection flips phase
Constructive interference occurs when $2n_{\text{film}}(2t)=(m+\tfrac{1}{2})\lambda_0$
Constructive interference depends only on incident intensity, not on $t$
Constructive interference requires $2n_{\text{film}}(2t)=m\lambda_0$ because phase shifts can be ignored
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.
A thin coating (index greater than air) is deposited on a lower-index plastic substrate. Light reflects from the air–coating and coating–plastic surfaces and recombines in air. For normal incidence, which condition produces constructive interference in reflected light?
$2n_{\text{coat}}t=m\lambda$
$2n_{\text{coat}}t=(m+\tfrac12)\lambda$
Constructive interference occurs when the reflected intensity is smallest
$t=\lambda$ (in air), independent of $n_{\text{coat}}$
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.
A film with refractive index between air and glass is illuminated normally by monochromatic light. Reflections from the top and bottom surfaces recombine in air. If reflected light is minimized at wavelength $\lambda$, which optical path condition in the film is satisfied?
No path condition; only the film’s reflectivity determines minima
$n_{\text{film}}t=\lambda$ only
$2n_{\text{film}}t=m\lambda$
$2n_{\text{film}}t=(m+\tfrac{1}{2})\lambda$
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: $2nt = (m + \tfrac{1}{2}) \lambda$.
Monochromatic light in air is normally incident on a thin film on a substrate. The film’s refractive index is greater than both air and the substrate. Reflections from the air–film surface and film–substrate surface return into air and interfere. The top reflection (low to high $n$) undergoes a $ rac{1}{2}\lambda$ phase shift, while the bottom reflection (high to low $n$) does not. The film thickness is $t$. Which condition produces constructive interference in the reflected light?
Constructive interference occurs only if the two reflected rays have equal intensities
$2n_{\text{film}}t=m\lambda_0$
Constructive interference occurs when the film is thicker than the wavelength in air, regardless of $t$
$2n_{\text{film}}t=(m+\tfrac{1}{2})\lambda_0$
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.
A thin film of thickness $t$ is between air (lower $n$) and a liquid (higher $n$ than the film). Monochromatic light in air strikes the film at normal incidence; reflections from the top and bottom surfaces recombine in air. The top reflection undergoes a $ rac{1}{2}\lambda$ phase shift (low to high $n$), and the bottom reflection also undergoes a $ rac{1}{2}\lambda$ phase shift (film to higher-$n$ liquid). With two phase reversals, which condition on $t$ minimizes the reflected light?
$2n_{\text{film}}t=m\lambda_0$
$2n_{\text{film}}t=(m+\tfrac{1}{2})\lambda_0$
Minimization occurs when the incident intensity is halved, independent of $t$
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.
A thin transparent coating (index higher than air but lower than the glass beneath it) is applied to a glass lens. Monochromatic light in air hits the coating at normal incidence. Reflections occur at the air–coating surface and the coating–glass surface, and the two reflected rays return into air. Because each reflection is from lower $n$ to higher $n$, both reflected rays undergo a $ rac{1}{2}\lambda$ phase shift. The coating thickness is $t$. Which condition on $t$ produces constructive interference in the reflected light?
$2n_{\text{coat}}t=(m+\tfrac{1}{2})\lambda_0$
Constructive interference occurs when the coating increases reflected intensity, not from thickness
Constructive interference occurs only if $n_{\text{coat}}=n_{\text{glass}}$, independent of $t$
$2n_{\text{coat}}t=m\lambda_0$
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.