Thin-film interference

One formula or one step of reasoning. Core exercises first, the rest is extra practice.

1. Exercise 14-1Core

   The refractive index of a soap film is $1.33$, and at one point its thickness is $420\,\text{nm}$. Light arrives at normal incidence on the film.

   What is the path difference between the waves reflected from the two surfaces of the film? Give your answer in nanometres.

   nmCheck
   Answer

   $1120\,\text{nm}$

   Show the solution

   Given

   refractive index of the soap film $n = 1.33$, thickness of the film $d = 420\,\text{nm}$, normal incidence

   Find

   the path difference $\Delta s$ between the waves reflected from the two surfaces, in nanometres

   Principle

   The wave reflected from the lower surface travels through the film and back, that is, an extra distance $2d$. In the film the wavelength is $n$ times shorter than in air, so the extra distance holds as many wavelengths as the distance $2nd$ does in air. At normal incidence the path difference is therefore $2nd$.

   Formula

   $\Delta s = 2nd$

   Conversions

   Not needed: the thickness is in nanometres, and the refractive index has no unit.

   Substitution

   $\Delta s$$= 2 \cdot 1.33 \cdot 420\,\text{nm}$

   $= 1117.2\,\text{nm}$

   Answer

   $1120\,\text{nm}$

   That is, $1.12\,\mu\text{m}$, three significant figures like the refractive index $1.33$ and the thickness $420\,\text{nm}$.

   Reasonableness

   The path difference is more than twice the thickness of the film, because the light travels through the film and back and the wavelength in the film is shorter. Without the refractive index the path difference would be $840\,\text{nm}$; the factor $1.33$ increases it by about a third, as in the result.

2. Exercise 14-2Core

   The wavelength of red laser light in air is $633\,\text{nm}$. What is the wavelength of the light in magnesium fluoride, whose refractive index is $1.38$?

   nmCheck
   Answer

   $459\,\text{nm}$

   Show the solution

   Given

   wavelength in air $\lambda = 633\,\text{nm}$, refractive index of magnesium fluoride $n = 1.38$

   Find

   the wavelength in magnesium fluoride $\lambda_n$ in nanometres

   Principle

   In a material light travels at the speed $v = c : n$, but its frequency $f$ does not change at an interface. Since $\lambda = v : f$, the wavelength shortens in the same ratio as the speed, that is, by the factor $n$.

   Formula

   $\lambda_n$$= \dfrac{v}{f}$

   $= \dfrac{c}{nf}$

   so $\lambda_n = \dfrac{\lambda}{n}$

   Conversions

   Not needed: the wavelength is in nanometres, and the refractive index has no unit.

   Substitution

   $\lambda_n$$= \dfrac{633\,\text{nm}}{1.38}$

   $= 458.70\,\text{nm}$

   Answer

   $459\,\text{nm}$

   Three significant figures like the wavelength $633\,\text{nm}$ and the refractive index $1.38$.

   Reasonableness

   The wavelength in the material is shorter than in air, because the refractive index is greater than one. The colour of the light still does not change, because the frequency stays the same. Check: $1.38 \cdot 458.70\,\text{nm} = 633\,\text{nm}$.

3. Exercise 14-3

   An $\text{MgF}_2$ coating ($n = 1.38$) is evaporated onto the glass of a camera lens so that light of wavelength $500\,\text{nm}$ is reflected as little as possible. The refractive index of the glass is $1.52$, and the light arrives at normal incidence.

   How thick is the thinnest such coating?

   nmCheck
   Answer

   $90.6\,\text{nm}$

   Show the solution

   Given

   refractive index of the coating $n = 1.38$, of the glass $1.52$, wavelength $\lambda = 500\,\text{nm}$, normal incidence

   Find

   the thickness $d$ of the thinnest reflection-reducing coating, in nanometres

   Principle

   Both reflections occur at a material of higher refractive index ($1.00 \to 1.38 \to 1.52$), so both waves are inverted, and the inversions cancel each other. The waves then weaken each other when $2nd = (k + \tfrac{1}{2})\lambda$, and the thinnest coating is given by $k = 0$.

   Formula

   $2nd = \tfrac{1}{2}\lambda$, so $d = \dfrac{\lambda}{4n}$

   Conversions

   Not needed: the wavelength and the thickness are both in nanometres.

   Substitution

   $d$$= \dfrac{500\,\text{nm}}{4 \cdot 1.38}$

   $= 90.580\,\text{nm}$

   Answer

   $90.6\,\text{nm}$

   Three significant figures like the refractive index $1.38$ and the wavelength $500\,\text{nm}$.

   Reasonableness

   The coating is a quarter of the wavelength in the coating, $500\,\text{nm} : 1.38 = 362\,\text{nm}$, and $362\,\text{nm} : 4 \approx 91\,\text{nm}$. The result is a little smaller than the $99.6\,\text{nm}$ coating of the chapter for a wavelength of $550\,\text{nm}$, because the wavelength is shorter.

4. Exercise 14-4

   The refractive index of a soap film is $1.33$, and there is air on both sides of it. Light arrives at normal incidence on the film.

   How thick is the thinnest film that reflects light of wavelength $600\,\text{nm}$ most strongly?

   nmCheck
   Answer

   $113\,\text{nm}$

   Show the solution

   Given

   refractive index of the soap film $n = 1.33$, air on both sides, wavelength $\lambda = 600\,\text{nm}$, normal incidence

   Find

   the thickness $d$ of the thinnest strongly reflecting film, in nanometres

   Principle

   The reflection from the upper surface, from air into the film, inverts the wave; the reflection from the lower surface, from the film into air, does not. Because of the single inversion, the waves strengthen each other when $2nd = (k + \tfrac{1}{2})\lambda$, and the thinnest such film is given by $k = 0$.

   Formula

   $2nd = \tfrac{1}{2}\lambda$, so $d = \dfrac{\lambda}{4n}$

   Conversions

   Not needed: the wavelength and the thickness are both in nanometres.

   Substitution

   $d$$= \dfrac{600\,\text{nm}}{4 \cdot 1.33}$

   $= 112.78\,\text{nm}$

   Answer

   $113\,\text{nm}$

   Three significant figures like the refractive index $1.33$ and the wavelength $600\,\text{nm}$.

   Reasonableness

   The same formula $d = \lambda : (4n)$ gives the strongest reflection in a soap film but the weakest in a coating, because the number of inverting reflections is different. The thickness is greater than the $103\,\text{nm}$ of the simulation for a wavelength of $550\,\text{nm}$, as it should be for a longer wavelength.

5. Exercise 14-5

   The refractive index of the plastic of a spectacle lens is $1.60$. At normal incidence a single-layer anti-reflection coating would cancel the reflection completely if the waves reflected from its two surfaces were equally strong.

   What should the refractive index of the coating then be?

   Check
   Answer

   $1.26$

   Show the solution

   Given

   refractive index of the plastic of the lens $n_\text{plastic} = 1.60$, normal incidence

   Find

   the refractive index $n_\text{c}$ of the ideal coating

   Principle

   The more the refractive indices at an interface differ, the stronger the reflected wave is. When the waves reflected from the upper and lower surfaces of the coating are equally strong, they can cancel each other completely. At normal incidence this happens when the refractive index of the coating is the square root of the refractive index of the lens material, as stated in the chapter: then $1.00 : n_\text{c} = n_\text{c} : n_\text{plastic}$.

   Formula

   $n_\text{c} = \sqrt{n_\text{plastic}}$

   Conversions

   Not needed: the refractive indices have no unit.

   Substitution

   $n_\text{c} = \sqrt{1.60} = 1.2649$

   Answer

   $1.26$

   Three significant figures like the refractive index $1.60$.

   Reasonableness

   The refractive index of the coating lies between $1.00$ for air and $1.60$ for the plastic, so both reflections invert the wave. The result is a little greater than the $1.23$ calculated for ordinary glass, because the refractive index of the plastic is greater. Check: $1.2649 \cdot 1.2649 = 1.60$.

Two steps, working with measured data, a unit conversion or rearranging a formula. Core exercises first, the rest is extra practice.

1. Exercise 14-6Core

   The thickness of a soap film ($n = 1.33$) is $240\,\text{nm}$ at one point, and there is air on both sides of it. White light arrives at normal incidence on the film.

   Which wavelength of visible light ($400$–$700\,\text{nm}$) is missing from the reflected light?

   nmCheck
   Answer

   $638\,\text{nm}$

   Show the solution

   Given

   refractive index of the soap film $n = 1.33$, thickness $d = 240\,\text{nm}$, air on both sides, visible light $400$–$700\,\text{nm}$

   Find

   the wavelength $\lambda$ missing from the reflected light, in nanometres

   Principle

   The reflection from the upper surface, from air into the film, inverts the wave; the reflection from the lower surface, from the film into air, does not. Because of the single inversion, the waves weaken each other when the path difference is a whole number of wavelengths, $2nd = k\lambda$. We calculate $\lambda$ for $k = 1, 2$ and choose the visible one.

   Formula

   $2nd = k\lambda$, so $\lambda = \dfrac{2nd}{k}$

   Conversions

   Not needed: the thickness and the wavelength are both in nanometres.

   Substitution

   $k = 1$:

   $\lambda$$= \dfrac{2 \cdot 1.33 \cdot 240\,\text{nm}}{1}$

   $= 638.4\,\text{nm}$

   for $k = 2$ we get $319.2\,\text{nm}$, which is ultraviolet.

   Answer

   $638\,\text{nm}$

   Three significant figures like the refractive index $1.33$ and the thickness $240\,\text{nm}$.

   Reasonableness

   The missing light is red, so the film looks blue-green at this point. The result is close to, and greater than, the $532\,\text{nm}$ missing for the $200\,\text{nm}$ film of the chapter, as it should be for a thicker film: the missing wavelength grows in the same ratio as the thickness. The thickness is $1.2$ times as great, and $1.2 \cdot 532\,\text{nm}$ is $638.4\,\text{nm}$.

2. Exercise 14-7Core

   A glass lens ($n = 1.52$) has a $110\,\text{nm}$ thick $\text{MgF}_2$ coating ($n = 1.38$) on its surface. Light arrives at normal incidence on the coating.

   For which wavelength of visible light ($400$–$700\,\text{nm}$) is the reflection from the coating weakest?

   nmCheck
   Answer

   $607\,\text{nm}$

   Show the solution

   Given

   refractive index of the coating $n = 1.38$, thickness $d = 110\,\text{nm}$, refractive index of the glass $1.52$, visible light $400$–$700\,\text{nm}$

   Find

   the most weakly reflected wavelength $\lambda$ in nanometres

   Principle

   Both reflections occur at a material of higher refractive index ($1.00 \to 1.38 \to 1.52$), so the inversions cancel each other. The waves weaken each other when $2nd = (k + \tfrac{1}{2})\lambda$. We calculate $\lambda$ for $k = 0, 1$ and choose the visible one.

   Formula

   $2nd = (k + \tfrac{1}{2})\lambda$, so

   $\lambda$$= \dfrac{2nd}{k + \frac{1}{2}}$

   Conversions

   Not needed: the thickness and the wavelength are both in nanometres.

   Substitution

   $k = 0$:

   $\lambda$$= \dfrac{2 \cdot 1.38 \cdot 110\,\text{nm}}{0.5}$

   $= \dfrac{303.6\,\text{nm}}{0.5}$

   $= 607.2\,\text{nm}$

   for $k = 1$ we get $303.6\,\text{nm} : 1.5 = 202.4\,\text{nm}$, which is ultraviolet.

   Answer

   $607\,\text{nm}$

   Three significant figures like the thickness $110\,\text{nm}$ and the refractive index $1.38$.

   Reasonableness

   For $k = 0$ the coating is a quarter-wave layer: $4 \cdot 1.38 \cdot 110\,\text{nm} = 607.2\,\text{nm}$. The coating is thicker than the $99.6\,\text{nm}$ designed for $550\,\text{nm}$, so it cancels a longer, orange wavelength.

3. Exercise 14-8

   An $\text{MgF}_2$ coating ($n = 1.38$) is made on glass ($n = 1.52$) so that it reflects light of wavelength $520\,\text{nm}$ as little as possible at normal incidence. The manufacturer wants a layer thicker than the thinnest possible.

   How thick is the second thinnest coating that reflects this wavelength most weakly?

   nmCheck
   Answer

   $283\,\text{nm}$

   Show the solution

   Given

   refractive index of the coating $n = 1.38$, of the glass $1.52$, wavelength $\lambda = 520\,\text{nm}$, normal incidence

   Find

   the thickness $d$ of the second thinnest destructive coating, in nanometres

   Principle

   Both reflections invert the wave ($1.00 \to 1.38 \to 1.52$), so the inversions cancel each other, and the waves weaken each other when $2nd = (k + \tfrac{1}{2})\lambda$. The thinnest coating is $k = 0$, so the second thinnest is given by $k = 1$.

   Formula

   $2nd = (1 + \tfrac{1}{2})\lambda$, so $d = \dfrac{3\lambda}{4n}$

   Conversions

   Not needed: the wavelength and the thickness are both in nanometres.

   Substitution

   $d$$= \dfrac{3 \cdot 520\,\text{nm}}{4 \cdot 1.38}$

   $= \dfrac{1560\,\text{nm}}{5.52}$

   $= 282.61\,\text{nm}$

   Answer

   $283\,\text{nm}$

   Three significant figures like the refractive index $1.38$ and the wavelength $520\,\text{nm}$.

   Reasonableness

   The thinnest coating would be $520\,\text{nm} : (4 \cdot 1.38) = 94.2\,\text{nm}$, and the second thinnest is three times that, because the path difference increases by one wavelength, that is, by two half wavelengths. Check:

   $2 \cdot 1.38 \cdot 282.61\,\text{nm}$

   $= 780\,\text{nm}$

   $= 1.5 \cdot 520\,\text{nm}$
4. Exercise 14-9

   In a beam splitter a thin titanium dioxide layer is made on glass ($n = 1.52$) so that it reflects light of wavelength $550\,\text{nm}$ as much as possible. Assume that the refractive index of the layer is $2.30$ and that the light arrives from air at normal incidence on the layer.

   How thick is the thinnest such layer?

   nmCheck
   Answer

   $59.8\,\text{nm}$

   Show the solution

   Given

   refractive index of the layer $n = 2.30$, of the glass $1.52$, wavelength $\lambda = 550\,\text{nm}$, light from air at normal incidence

   Find

   the thickness $d$ of the thinnest strongly reflecting layer, in nanometres

   Principle

   The reflection from the upper surface, from air into the layer ($1.00 \to 2.30$), inverts the wave. At the lower surface the light is reflected from the layer into the glass ($2.30 \to 1.52$), that is, at a material of lower refractive index, so the wave is not inverted. Because of the single inversion, the conditions are the same as in a soap film: the waves strengthen each other when $2nd = (k + \tfrac{1}{2})\lambda$, and the thinnest layer is given by $k = 0$.

   Formula

   $2nd = \tfrac{1}{2}\lambda$, so $d = \dfrac{\lambda}{4n}$

   Conversions

   Not needed: the wavelength and the thickness are both in nanometres.

   Substitution

   $d$$= \dfrac{550\,\text{nm}}{4 \cdot 2.30}$

   $= 59.783\,\text{nm}$

   Answer

   $59.8\,\text{nm}$

   Three significant figures like the refractive index $2.30$ and the wavelength $550\,\text{nm}$.

   Reasonableness

   The layer is thinner than an $\text{MgF}_2$ coating of $99.6\,\text{nm}$ for the same wavelength, because the refractive index is greater and the wavelength in the layer is therefore shorter. The decisive point is that behind the lower surface there is a lower refractive index: the same quarter-wave thickness therefore strengthens the reflection instead of weakening it.

Optional: application, two ideas together or a new situation where you reason for yourself.

1. Exercise 14-10

   A soap film ($n = 1.33$) made in a vertical wire frame is lit with the yellow light of a sodium lamp, with wavelength $589\,\text{nm}$, arriving at normal incidence on the film.

   Gravity thins the film at the top, and the topmost part of the film looks black. Below it bright and dark horizontal bands alternate, and the film gets thicker downwards.

   How thick is the film at the second dark band, when the topmost black region is not counted as a band?

   nmCheck
   Answer

   $443\,\text{nm}$

   Show the solution

   Given

   refractive index of the soap film $n = 1.33$, air on both sides, wavelength $\lambda = 589\,\text{nm}$, normal incidence; the film thickens from top to bottom, and the topmost part is black

   Find

   the thickness $d$ of the film at the second dark band, in nanometres

   Principle

   The reflection from the upper surface inverts the wave and that from the lower surface does not, so the waves weaken each other when $2nd = k\lambda$. The black top is the value $k = 0$, where the film is much thinner than the wavelength. The film thickens downwards, so the next dark bands are in order $k = 1$ and $k = 2$; at the second dark band $k = 2$. The bright bands lie between the dark ones.

   Formula

   $2nd = 2\lambda$, so $d = \dfrac{\lambda}{n}$

   Conversions

   Not needed: the wavelength and the thickness are both in nanometres.

   Substitution

   $d$$= \dfrac{589\,\text{nm}}{1.33}$

   $= 442.86\,\text{nm}$

   Answer

   $443\,\text{nm}$

   Three significant figures like the refractive index $1.33$ and the wavelength $589\,\text{nm}$.

   Reasonableness

   The first dark band is at a thickness of $589\,\text{nm} : (2 \cdot 1.33) = 221\,\text{nm}$, and the second is twice that, because between consecutive dark bands the thickness always grows by the same amount, $\lambda : (2n)$. The thickness is less than a micrometre, as is fitting for a thin film.

2. Exercise 14-11

   A camera objective has four separate glass lenses, that is, eight surfaces at which light passes from air into glass or from glass into air. Each surface is coated so that it reflects $1.3\,\%$ of the light arriving at it at normal incidence.

   Assume that the glass does not absorb light and that light reflected once no longer reaches the sensor.

   What percentage of the light entering the objective gets through it?

   %Check
   Answer

   $90\,\%$

   Show the solution

   Given

   eight surfaces, each of which reflects $1.3\,\%$ of the light arriving at it; the glass does not absorb light, and reflected light does not return to the sensor

   Find

   the percentage of light that gets through the objective

   Principle

   Light does not disappear in the coating: what is not reflected goes through the surface. Through each surface therefore passes $98.7\,\%$ of the light arriving at it, since $100\,\% - 1.3\,\% = 98.7\,\%$. The next surface receives only the light that passed the previous one, so the fractions are multiplied together eight times.

   Formula

   $\dfrac{I_\text{out}}{I_0} = (1 - R)^8$

   Conversions

   $R = 1.3\,\% = 0.013$. Finally the fraction is multiplied by one hundred to give a percentage.

   Substitution

   $\dfrac{I_\text{out}}{I_0}$$= (1 - 0.013)^8$

   $= 0.987^8$

   $= 0.9006$

   $= 90.06\,\%$

   Answer

   $90\,\%$

   Two significant figures like the reflected fraction $1.3\,\%$.

   Reasonableness

   Estimate: the eight surfaces reflect altogether about $8 \cdot 1.3\,\% \approx 10\,\%$, so about $90\,\%$ gets through. Without coatings each surface would reflect about $4.26\,\%$, and only $(1 - 0.0426)^8 \approx 70.6\,\%$ would get through. The coatings thus increase the light reaching the sensor by about $19$ percentage points, that is, by a good quarter ($90.06 : 70.6 \approx 1.28$), and reduce ghost images.

3. Exercise 14-12

   A soap film ($n = 1.33$) is in air, and white light arrives at normal incidence on the film. In the visible range the film reflects most strongly the wavelengths $630\,\text{nm}$ and $450\,\text{nm}$, and there are no other strongly reflected wavelengths between them.

   How thick is the film?

   nmCheck
   Answer

   $592\,\text{nm}$

   Show the solution

   Given

   refractive index of the soap film $n = 1.33$, air on both sides, most strongly reflected wavelengths $\lambda_1 = 630\,\text{nm}$ and $\lambda_2 = 450\,\text{nm}$, none between them

   Find

   the thickness $d$ of the film in nanometres

   Principle

   There is one inverting reflection, so the waves strengthen each other when $2nd = (k + \tfrac{1}{2})\lambda$. The same thickness strengthens both wavelengths. The shorter wavelength fits into the path difference more times, and since there are no others between them, its order is greater by one: $(k + \tfrac{1}{2})\lambda_1 = (k + \tfrac{3}{2})\lambda_2$. From this we first solve $k$ and then $d$.

   Formula

   $(k + \tfrac{1}{2})\lambda_1 = (k + \tfrac{3}{2})\lambda_2$, so

   $k$$= \dfrac{\tfrac{3}{2}\lambda_2 - \tfrac{1}{2}\lambda_1}{\lambda_1 - \lambda_2}$

   thickness

   $d$$= \dfrac{(k + \frac{1}{2})\lambda_1}{2n}$

   Conversions

   Not needed: the wavelengths and the thickness are all in nanometres.

   Substitution

   $k$$= \dfrac{1.5 \cdot 450\,\text{nm} - 0.5 \cdot 630\,\text{nm}}{630\,\text{nm} - 450\,\text{nm}}$

   $= \dfrac{360\,\text{nm}}{180\,\text{nm}}$

   $= 2$

   so

   $d$$= \dfrac{2.5 \cdot 630\,\text{nm}}{2 \cdot 1.33}$

   $= \dfrac{1575\,\text{nm}}{2.66}$

   $= 592.11\,\text{nm}$

   Answer

   $592\,\text{nm}$

   Three significant figures like the refractive index $1.33$ and the wavelengths.

   Reasonableness

   Check: $3.5 \cdot 450\,\text{nm} = 1575\,\text{nm}$ and $2.5 \cdot 630\,\text{nm} = 1575\,\text{nm}$, so the same path difference strengthens both. The other orders give $1575\,\text{nm} : 1.5 = 1050\,\text{nm}$ (infrared) and $1575\,\text{nm} : 4.5 = 350\,\text{nm}$ (ultraviolet), so in the visible range there really are only these two.