Questions: 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 11 18

Interference and Diffraction Solutions

1. Young's experiment is performed with sodium light (λ = 589 nm). Fringes are measured carefully on a screen 1.50 m away from the double slit, and the centre of the twentieth bright fringe (not counting the central bright fringe) is found to be 11.9 mm from the centre of the central bright fringe. What is the separation of the two slits?

   The equation for obtaining bright fringes (i.e. constructive interference) in Young's experiment is

   dsin(θ)= mλ ,

   where m is the order of the fringe and d is the slit separation. The central fringe is m = 0, so the twentieth fringe not including the centre indicates that m = 20.

   For small angles, sin(θ) ≈ y/S. Solving for d, we find

   d = mλS/y = (20)(589 × 10^-9 m)(1.50 m)/(0.0119 m) = 1.48 mm .



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2. You repeat the above experiment with light of wavelength λ = 502 nm. What does the slit separation have to be for the fifteenth (not counting the centre) to fall at 11.9 mm? Again the screen is 1.50 m away from the double slit.

   In this problem the position of the fringe is the same in each case so θ and sin(θ) are the same. In the initial case

   dsin(θ) = 20λ

   and in the second case

   d' sin(θ) = 15 λ' .

   Eliminating the common factor sin(θ) and rearranging for d', yields

   d' = d (15λ' / 20λ) = (1.48 mm)(15)(502)/(20)(589) = 0.95 mm.



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3. The diagram below shows the paths of two rays that meet at a screen. The light has wavelength λ. The upper ray travels straight to the screen. The other ray is diverted by four mirrors as shown. The vertical separation of the two rays is exaggerated for clarity. Derive the equation that determines if the two rays interfere constructively at the screen

   

   
   

   We concentrate on what is different between the two rays. The bottom ray is reflected four times. That represents either four flip reflections which is no net change or four non-flip reflections which is also no net change. So the mirrors can be ignored.

   The bottom ray travels 2t more that the upper ray. So the number of extra wavelengths in the bottom ray compared to the top is 2t/λ. We will get constructive interference if

   2t /λ = n, where n is an integer.

   

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4. The diagram below shows the paths of two rays that meet at a screen. The light has wavelength λ. The upper ray passes though a piece of glass of index n and thickness t. The other ray is diverted by four mirrors as shown. The vertical separation of the two rays is exaggerated for clarity. Derive the equation that determines if the two rays interfere constructively at thescreen.

   

   We concentrate on what is different between the two rays. The bottom ray is reflected four times. That represents either four flip reflections which is no net change or four non-flip reflections which is also no net change. So the mirrors can be ignored.

   The bottom ray travels 3t in air while the upper ray travels t in glass. So the number of extra wavelengths in the bottom ray compared to the top is 3t/λ – t/λ_glass. We will get constructive interference if

   (3t/λ – t/λ_glass) = m, where m is an integer.

   Recalling λ_glass = λ/n, our equation reduces to

   (t/λ)(3 – n) = m, where m is an integer.



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5. What is the thinnest film of a 1.40 refractive index coating on glass (n = 1.50) for which destructive interference of the green component (500 nm) of an incident white light beam in air can take place by reflection?
   

   Note that the angles in the sketch are exaggerated for clarity. Interference is governed by phase differences caused by reflection and by differing path lengths.
   

   First consider the phase difference caused by reflection. Ray 1 is reflected at an air/film interface. Since n_air < n_film, the phase of Ray 1 is flipped relative to the incident ray. Ray 2 is reflected at an film/glass interface. Since n_film < n_glass, the phase of Ray 2 is also flipped relative to the incident ray. Since both rays have flipped, Δ_reflection/2π = 0.

   Next consider the phase difference caused by the fact that Ray 2 travels an extra distance 2t (remember that we assume that the angles involved are small). This corresponds to a phase difference (number of wavelengths) of Δ_travelled/2π = 2t/λ_film, where λ_film is the wavelength of the light in the thin film. The wavelength in the film is related to the wavelength in air by n_filmλ_film = n_airλ_air.

   Since Δ_reflection/2π = 0, we get destructive interference when
   

   Δ_travelled/2π = ½n,        where n = 1, 3, 5, … .

   Substituting in our expression for Δ_travelled we have

   2t / (n_airλ_air/n_film) = ½n.

   Solving for t yields,
   

   t = ¼n(n_airλ_air/n_film) ,        where n = 1, 3, 5, … .

   The thinnest film for this to occur is when n = 1, so
   

   t = ¼(1)(1.00)(500 nm)/(1.40) = 89.3 nm .

   

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6. A layer of oil of thickness 200 nm floats on top of a layer of water of thickness 400 nm resting on a flat metallic mirror. The index of refraction of the oil is 1.24 and that of the water is 1.33. A beam of light is normally incident on these layers. What are the visible-light wavelength(s) of the beam if the light reflected by the top surface of the oil interferes destructively with the light reflected by the mirror?
   

   Note that the angles in the sketch are exaggerated for clarity. Interference is governed by phase differences caused by reflection and by differing path lengths.
   

   First consider the phase difference caused by reflection. Ray 1 is reflected at an air/oil interface. Since n_air < n_oil, the phase of Ray 1 is flipped relative to the incident ray. Ray 2 is reflected at an water/mirror interface. We are not told the index of refraction of the metal mirror, however since it is not transparent n must be infinite. Since n_water < n_metal, the phase of Ray 2 is flipped relative to the incident ray. Since both rays have flipped, Δ_reflection/2π = 0.
   

   Next consider the phase difference caused by the fact that Ray 2 travels an extra distance in each thin film, 2t_oil and 2t_water(remember that we assume that the angles involved are small). This corresponds to a phase difference (number of wavelengths) of
   

   Δ_travelled/2π = 2t_oil/λ_oil + 2t_water/λ_water ,

   where λ is the wavelength of the light in the particular medium. The wavelength in the oil film is related to the wavelength in air by n_oilλ_oil = n_airλ_air. The wavelength in the water film is related to the wavelength in air by n_waterλ_water = n_airλ_air.
   

   Since Δ_reflection/2π = 0, we get destructive interference when
   

   Δ_travelled/2π = ½n,        where n = 1, 3, 5, … .

   Substituting in our expression for Δ_travelled we have
   

   2t_oil / (n_airλ_air/n_oil) + 2t_water / (n_airλ_air/n_water) = ½n .

   Solving for λ_air yields,
   

   λair	= (2/n)[ 2toil / (nair/noil) + 2twater / (nair/nwater)].
   	= 4[noiltoil + nwatertwater]/n
   	= 4[(1.24)(200 nm) + (1.33)(400 nm)]/n
   	= 3120 nm / n

   Next we examine the odd value of n and find which wavelengths fall in the visible range

   n	λair = 3120/n (nm)
   1	3120
   3	1040
   5	624
   7	446
   9	347
   11	284
   13	240

   Only visible light with wavelengths of 624 nm and 446 nm would interefere destructively with this double film. All other wavelengths are outside the visible range of 400 nm < λ_visible < 700 nm.

   

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7. Two flat 10 cm long glass plates are shown in the diagram below. At one end of the plates are in contact, while at the other they are separated by a distance of 0.020 mm. Incident light (λ = 500 nm) shines normally onto the glass and is reflected of the bottom layer of the top plate and the top layer of the of the bottom plate. What is the spacing of the interference fringes? Is the fringe adjacent to the line of contact bright or dark?
   

   Note that the angles in the sketch are exaggerated for clarity. Interference is governed by phase differences caused by reflection and by differing path lengths.
   

   First consider the phase difference caused by reflection. Ray 1 is reflected at an glass/air interface. Since n_glass > n_air, the phase of Ray 1 is NOT flipped relative to the incident ray. Ray 2 is reflected at an air/glass interface. Since n_air < n_glass, the phase of Ray 2 is flipped relative to the incident ray. Since only one ray has flipped, Δ_reflection/2π = ½.
   

   Next consider the phase difference caused by the fact that Ray 2 travels an extra distance through air of 2d (remember that we assume that the angles involved are small). This corresponds to a phase difference (number of wavelengths) of
   

   Δ_travelled/2π = 2d/λ_air,

   where λ_air is the wavelength of the light in air.
   

   We are asked to find where the interference fringes are, but the problem doesn't specify whether we are interested in bright or dark fringes, so we will do both.
   

   Since Δ_reflection/2π = ½, we get destructive interference and dark fringes when
   

   Δ_travelled/2π = n,        where n = 0, 1, 2, 3, … .

   Substituting in our expression for Δ_travelled we have
   

   2d / λ_air = n.           (1)

   The separation d varies with x according to the relationship,
   

   d = (h/l)x ,           (2)

   which may be derived from simple geometry. Combining equations (1) and (2) and solving for x we find,
   

   x = nl_air /2h = n (0.10 m)(500 nm) / (2)(0.020 × 10^-3 m) = n (1.25 mm) ,           (3)

   where n is 0, 1, 2, 3, …
   

   Thus the distance between two consecutive dark fringes is just 1.25 mm since n increases in increments of 1.
   

   Since Δ_reflection/2π = ½, we get constructive interference and bright fringes when
   

   Δ_travelled/2π = ½n,        where n = 1, 3, 5, … .

   Substituting in our expression for Δ_travelled we have
   

   2d / λ_air = ½n.           (4)

   Combining equations (1) and (3) and solving for x we find,
   

   x = nl_air / 4h = n (0.10 m)(500 nm) / (4)(0.020 × 10^-3 m) = n (0.625 mm) ,           (5)

   where n is 1, 3, 5, …
   

   Thus the distance between two consecutive bright fringes is just 2 × 0.625 mm = 1.25 mm, since n increases in increments of 2. Note that we of course get the same answer as for dark fringes.
   

   The line of contact occurs when x = 0. Examining equations (3) and (5), we see that n must be zero if x = 0. But n = 0 is only allowed for dark fringes. The line of contact is therefore a dark fringe.
   

   

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8. If in the above problem, the space between the plates is filled with transparent silicone grease (n = 1.5), the top plate has refractive index, n = 1.4, and the lower n = 1.6, what is the spacing of the dark fringes? Is the fringe at the line of contact bright or dark?

   Interference is governed by phase differences caused by reflection and by differing path lengths.
   

   First consider the phase difference caused by reflection. Ray 1 is reflected at an glass/grease interface. Since n_glass < n_grease, the phase of Ray 1 is flipped relative to the incident ray. Ray 2 is reflected at an grease/glass interface. Since n_grease < n_glass, the phase of Ray 2 is flipped relative to the incident ray. Since both rays have flipped, Δ_reflection/2π = 0.
   

   Next consider the phase difference caused by the fact that Ray 2 travels an extra distance through grease of 2d (remember that we assume that the angles involved are small). This corresponds to a phase difference (number of wavelengths) of
   

   Δ_travelled/2π = 2d/λ_grease,

   where λ_grease is the wavelength of the light in silicone grease. The wavelength in the grease is related to the wavelength in air by n_greaseλ_grease = n_airλ_air.

   Since Δ_reflection/2π = 0, we get destructive interference and dark fringes when
   

   Δ_travelled/2π = ½n,        where n = 1, 3, 5, … .

   Substituting in our expression for Δ_travelled we have
   

   2d / (n_airλ/n_grease) = ½n.            (1)

   The separation d varies with x according to the relationship,
   

   d = (h/l)x ,            (2)

   which may be derived from simple geometry. Combining equations (1) and (2) and solving for x we find,
   

   x	= nlnairλair /4hngrease
   	= n (0.10 m)(1)(500 nm) / (4)(0.020 × 10-3 m)(1.50)
   	= n (0.417 mm)

   where n is 1, 3, 5, …
   

   Thus the distance between two consecutive dark fringes is just 2 × 0.417 mm = 0.833 mm since n increases in increments of 2.
   

   Since Δ_reflection/2π = 0, we get constructive interference and bright fringes when
   

   Δ_travelled/2π = n,        where n = 0, 1, 2, 3, … .

   Substituting in our expression for Δ_travelled we have
   

   2d / (n_airλ/n_grease) = n.           (3)

   Combining equations (1) and (3) and solving for x we find,
   

   x	= nlnairλair / 4hngrease
   	= n (0.10 m)(500 nm) / (2)(0.020 × 10-3 m)
   	= n (0.833 mm)

   where n is 0,1, 2, 3, …
   

   Thus the distance between two consecutive bright fringes is just 0.833 mm, since n increases in increments of 1. Note that we of course get the same answer as for dark fringes.
   

   The line of contact occurs when x = 0. Examining equations for the location of the bright and dark fringes, we see that n must be zero if x = 0. But n = 0 is only allowed for bright fringes. The line of contact is therefore a bright fringe.
   

   

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9. A thin film of thickness 0.450 mm and index of refraction n = 1.50 is placed in one arm of a Michelson Interferometer. Light of wavelength λ = 550 nm is used. How many fringes does the pattern shift? In the entire setup was immersed in a fluid of index n = 1.20, how many fringes would the pattern shift?

   A fringe shift is when a bright fringe shifts to dark and then back to bright ( or vice versa), a total phase change of one wavelength.

   A fringe shift occurs when there is a path difference between the two beams of light which travel the different arms of the interferometer.
   

   (a) One beam travels twice through the film of thickness t. The film thus corresponds to 2t/λ_film wavelengths. The other beam travels through an equal amount of air, but because of the different index this corresponds to 2t/λ_air wavelengths. The difference in these is the cause of the shift in fringes. The number of fringes shifted is

   m = | 2t/λ_film - 2t/λ_air | .

   The absolute signs ensure that m is positive. We know that n_filmλ_film = n_airλ_air, so we get

   m	= (2t/λair) (nfilm/nair - 1)
   	= [2(0.450 × 10-3 m)/(550 × 10-9 m)](1.50/1.00 - 1)
   	= 818

   A shift of 818 fringes is observed.
   

   (b) When the interferometer is immersed in fluid, the number of fringes will be

   m = | 2t/λ_film - 2t/_fluid | .

   We know that n_filmλ_film = n_fluidλ_fluid = n_airλ_air, so we get
   

   m	= (2t/λair) (nfilm - nfluid)/nair
   	= [2(0.450 × 10-3 m)/(550 × 10-9 m)](1.50 - 1.20)/(1.00)
   	= 491

   A shift of 491 fringes is observed.
   

   

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10. An airtight chamber 5.0 cm long with glass windows is placed in one arm of a Michelson Interferometer. Light of wavelength 500 nm is used. Air is slowly evacuated from the chamber using a vacuum pump. This causes a shift of 60 fringes. Find the index of refraction of air at atmospheric pressure and room temperature. A fringe shift is when a point goes from dark to light to dark or vice versa.
    

    A fringe shift is when a bright fringe shifts to dark and then back to bright ( or vice versa), a total phase change of one wavelength.

    A fringe shift occurs when there is a path difference between the two beams of light which travel the different arms of the interferometer. Clearly the path difference here is 60 wavelengths.
    

    One beam travels twice through the vacuum of thickness t. The film thus corresponds to 2t/λ_vacuum wavelengths. The other beam travels through an equal amount of air, but because of the different index this corresponds to 2t/λ_air wavelengths. The difference in these is the cause of the observed shift in fringes. The number of fringes shifted is
    

    m = | 2t/λ_vacuum - 2t/λ_air | .

    The absolute signs ensure that m is positive. We know that n_airλ_air = λ_vacuum, so we get
    

    m = (2t/λ_vacuum) (n_air - 1) .

    Solving for n_air, we get
    

    n_air = 1 + mλ_vacuum/2t = 1 + (60)(500 × 10^-9 m) / (2)(0.05 m) = 1.0030 .




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11. In Young's double slit experiment, when a thin film of transparent material is placed over one of the slits, the central bright fringe of the white-light fringe system is displace by 3.6 fringes. The refractive index of the material is 1.40, and the wavelength of the light is 550 nm. (a) By how much does the film increase the optical path? (b) What is the exact thickness of the film?

    The m = 0 fringe occurs when there is no path difference between the light from each slit. The m = 1 fringe represents a difference of one wavelength and so on. In this problem the presence of the film represents a shift of n = 3.6 wavelengths.

    The light travels through the film only once on its way to the screen, it therefore contains t/λ_film wavelengths, where t is the thickness of the film (remember we assume small angles for the path). The corresponding number of wavelengths for a similar path of air is t/λ_air. The path difference therefore is
    

    n = |t/λ_film - t/λ_air| ,

    where n_filmλ_film = n_airλ_air, and the absolute signs ensure a positive value for n. Solving for t we find
    

    t = nλ_airn_air / (n_film - n_air) = ( 3.6)(550 nm)(1.00)/(1.40 - 1.00) = 4.95 × 10^-6 m .

    

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12. A 10-slit grating with d = 0.015 mm is 2.00 m from a screen. Light of wavelengths 550 nm and 575 nm shines though the slits. Where on the screen are the n = 3 bright fringes?

    Multi-slit bright fringes obey the same formula as the double slit,

    dsin(θ) = 3λ.

    Solving for θ we find θ = 0.1102 radians or (6.315°). This is not a small angle so using some trig (see figure below) to find the distance y that the fringe is from the centre, we find tan(θ) = y/L, so y = 2.00 tan(6.315°) = 0.221 m.

    

    
    



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13. On a flat screen 2.0 m from a single slit, the distance between the first-order minima is 13.3 cm for light of wavelength λ = 550 nm. What is the width of the slit? For light of wavelength λ = 600 nm, how far apart are the first-order minima?
    

    The diffraction minima of a single slit are governed by the equation

    asin(θ) = mλ ,

    where a is the width of the slit and m is the order of the minimum.

    (a) For first order minima, m = 1. Using the small angle approximation, sin(θ) ≈ y/S, we find

    a = λS/y = (600 × 10^-9 m)(2.0 m)/(½ 0.133 m) = 1.65 × 10^-5 m .           (1)

    The slit is 0.017 mm wide.

    (b) We can rearrange equation (1) to solve for d

    d = 2y_new = 2λ_newS/a = 2y_oldλ_new/λ_old = (13.3 cm)(600 nm)/(550 nm) = 14.5 cm .

    The first order minima are 14.5 cm apart.



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14. You observe a double slit pattern using slit separation d = 0.011 mm. There are only 21 fringes in the central diffraction maximum. What is the slit width D?

    21 fringes means that the n = 0 to n = 10 interference fringes are visible in the central maximum of the diffraction pattern. The first diffraction minima must be obscuring the n = 11 interference fringe. Assuming the diffraction minima is at the same location, i.e. θ, we have

    Dsin(θ) = λ      1^st diffraction minima

    and

    dsin(θ) = 11λ  interference maxima.

    Eliminating the common factor sin(θ), we find

    D = d / 11 = 0.0010 mm.



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15. It is claimed that some professional baseball players can see which way the ball is spinning as it travels toward home plate. One way to judge this is to estimate the distance at which a batter can first hope to resolve two points on opposite sides of a baseball, which has a diameter of 0.0738 m. (a) Estimate this distance, assuming the pupil of the eye has a diameter of 2.0 mm, the material within the eye has a refractive index of 1.36, and the wavelength of light is 550 nm in a vacuum. (b) Considering that the distance between the pitcher's mound and home plate is 18.4 m, can you rule out such a claim?

    Two points may be resolved, see as two distinct points, when the angular separation is greater than

    θ_minimum = 1.22λ_medium/D ,

    where D is the diameter of the aperature and λ_medium is the wavelength of light in the medium where the light detector is. Note that D is the diameter of the pupil in this case and that the detector is the retina which is immersed in the vitreous humour of the eye. As well n_eyeλ_eye = n_airλ_air. Thus

    θminimum	= 1.22nairλair / neyeD
    	= (1.22)(1.00)(550 × 10-9 m)/(1.36)(0.002 m)
    	= 2.47 × 10-4 radians

    Now y ≈ Lθ, so

    L_maximum = y / θ_minimum = 0.0738 m / 2.47 × 10^-4 = 299 m .

    So the batter can resolve the edges of the ball out to about 300 m.

    Since the pitcher's mound is only 18.4 m away from the batter, the claim may be true.



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16. Ignoring atmospheric effects, how far apart must two objects be on the moon before the eye can resolve them? The distance to the moon's surface is 3.77 × 10⁸ m. Assume λ = 550 nm and n = 1.34. Assume that the diameter of the eye's pupil is 2.00 mm.
    

    Two points may be resolved, see as two distinct points, when the angular separation is greater than
    

    θ_minimum = 1.22λ_medium/D ,

    where D is the diameter of the aperature and λ_medium is the wavelength of light in the medium where the light detector is. Note that D is the diameter of the pupil in this case and that the detector is the retina which is immersed in the vitreous humour of the eye. As well n_eyeλ_eye = n_airλ_air. Thus
    

    θminimum	= 1.22nairλair/neyeD
    	= (1.22)(1.00)(550 × 10-9 m)/(1.34)(0.002 m)
    	= 2.50 × 10-4 radians

    Now y ≈ Lθ, so
    y_minimum = L_minimum = (3.77 × 10⁸ m)(2.50 × 10^-4) = 94 km .

    So objects about 100 km apart could be resolved as separate object. They would of course have to be large objects such as mountains or the like.



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17. Sodium light with wavelengths 588.99 nm and 589.59 nm is incident on a grating with 5500 lines per centimetre. A screen is placed 3.0 m beyond the grating. What is the distance between the two spectral lines in the first-order spectrum on the screen? In the second order spectrum?
    

    The spectral line are interference maxima and are thus covered by the equation
    

    dsin(θ) = mλ ,            (1)

    where is the particular wavelength of interest and d is the separation of the slits.
    

    We find the separation from
    

    d = 1 cm / 5500 = 1.818 × 10^-6 m .

    Let's evaluate θ₁ to see how big the angles are.
    

    θ₁ = arcsin(md/λ) = arcsin(1.818 m / 588.99 nm) = 0.3299321 radians = 18° .

    This is not a small angle, so we cannot use the small angle approximation directly, however Δy₁ ≈ R(θ₂ - θ₁) and Δy₂ ≈ R(θ₄ - θ₄) .
    

    Thus we find
    

    Δy₁ ≈ R(θ₂ - θ₁) = (3.0 m)(0.3302789 - 0.3299301) = 1.05 mm,

    and

    Δy₂ ≈ R(θ₄ - θ₃) = (3.0 m)(0.705760 - 0.704895) = 2.60 mm.

    

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18. A 10-slit grating with d = 0.015 mm is 2.00 m from a screen. Light of wavelengths 550 nm though the slits. What is the width of fringes? What are the nearest wavelength(s) of light that could also be present and resolved in second order? In third order?

    The angular width of the fringes are given by Δθ = λ / Nd. We want the linear width however. We recall that ΔS = RΔθ, so

    ΔS = Rλ / Nd = (2.00 m)(550 × 10^-9) / (10)(0.015 × 10^-3) = 0.0073 m = 7.3 mm.

    The wavelength resolution is given by Δλ/λ = 1 / nN. So in second order

    Δλ = λ / nN = 550 nm / 20 = 27.5 nm.

    That means we can resolve wavelengths below 550 – 27.5 nm = 522.5 nm or above 550 + 27.5 nm = 577.5 nm in second order.

    In third order

    Δλ = λ / nN = 550 nm / 30 = 18.3 nm.

    That means we can resolve wavelengths below 531.7 nm or above 568.3 nm in third order.

Questions? mike.coombes@kwantlen.ca