Chapters 36 & 37 Interference and Diffraction

Combination of Waves In general, when we combine two waves to form a composite wave, the composite wave is the algebraic sum of the two original waves, point by point in space [Superposition Principle]. When we add the two waves we need to take into account their: Direction Amplitude Phase + =

Combination of Waves The combining of two waves to form a composite wave is called: Interference The interference is constructive if the waves reinforce each other. + = Constructive interference (Waves almost in phase)

Combination of Waves The combining of two waves to form a composite wave is called: Interference The interference is destructive if the waves tend to cancel each other. + = (Close to  out of phase) (Waves almost cancel.) Destructive interference

Interference of Waves + = Constructive interference (In phase) + = (  out of phase) (Waves cancel) Destructive interference

Interference of Waves When light waves travel different paths, and are then recombined, they interfere. Each wave has an electric field whose amplitude goes like: E(s,t) = E 0 sin(ks-  t) î Here s measures the distance traveled along each wave’s path. Mirror 1 2 * + = Constructive interference results when light paths differ by an integer multiple of the wavelength:  s = m

Interference of Waves When light waves travel different paths, and are then recombined, they interfere. Each wave has an electric field whose amplitude goes like: E(s,t) = E 0 sin(ks-  t) î Here s measures the distance traveled along each wave’s path. Mirror 1 2 * Destructive interference results when light paths differ by an odd multiple of a half wavelength:  s = (2m+1) /2 + =

Interference of Waves Coherence: Most light will only have interference for small optical path differences (a few wavelengths), because the phase is not well defined over a long distance. That’s because most light comes in many short bursts strung together. Incoherent light: (light bulb) random phase “jumps”

Interference of Waves Coherence: Most light will only have interference for small optical path differences (a few wavelengths), because the phase is not well defined over a long distance. That’s because most light comes in many short bursts strung together. Incoherent light: (light bulb) Laser light is an exception: Coherent Light: (laser) random phase “jumps”

Thin Film Interference We have all seen the effect of colored reflections from thin oil films, or from soap bubbles. Film; e.g. oil on water

Thin Film Interference We have all seen the effect of colored reflections from thin oil films, or from soap bubbles. Film; e.g. oil on water Rays reflected off the lower surface travel a longer optical path than rays reflected off upper surface.

Thin Film Interference We have all seen the effect of colored reflections from thin oil films, or from soap bubbles. Film; e.g. oil on water Rays reflected off the lower surface travel a longer optical path than rays reflected off upper surface. If the optical paths differ by a multiple of, the reflected waves add. If the paths cause a phase difference , reflected waves cancel out.

Thin Film Interference 1 t 2 oil on water optical film on glass soap bubble n = 1 n > 1 Ray 1 has a phase change of  upon reflection (1<n) Ray 2 travels an extra distance 2t (normal incidence approximation) Constructive interference: rays 1 and 2 are in phase  2 t = (m+1/2) n  2 n t = (m + 1/2) [ n = /n] Destructive interference: rays 1 and 2 are  out of phase  2 t = m n  2 n t = m

Thin Film Interference Thin films work with even low coherence light, as paths are short Different wavelengths will tend to add constructively at different angles, and we see bands of different colors. When ray 2 is in phase with ray 1, they add up constructively and we see a bright region. When ray 2 is  out of phase, the rays interfere destructively. This is how anti-reflection coatings work. 1 t 2 oil on water optical film on glass soap bubble n = 1 n > 1

Diffraction What happens when a planar wavefront of light interacts with an aperture? If the aperture is large compared to the wavelength you would expect this.... …Light propagating in a straight path.

If the aperture is small compared to the wavelength would you expect this? Diffraction Not really…

In fact, what happens is that: a spherical wave propagates out from the aperture. All waves behave this way. Diffraction This phenomenon of light spreading in a broad pattern, instead of following a straight path, is called: DIFFRACTION If the aperture is small compared to the wavelength, would you expect the same straight propagation? … Not really

Angular Spread:  ~ /a Slit width a:  I (Actual intensity distribution) Diffraction

Huygen’s Principle Huygen first explained this in 1678 by proposing that all planar wavefronts are made up of lots of spherical wavefronts..

Huygen’s Principle Huygen first explained this in 1678 by proposing that all planar wavefronts are made up of lots of spherical wavefronts.. That is, you see how light propagates by breaking a wavefront into little bits

Huygen’s Principle Huygen first explained this in 1678 by proposing that all planar wavefronts are made up of lots of spherical wavefronts.. That is, you see how light propagates by breaking a wavefront into little bits, and then draw a spherical wave emanating outward from each little bit.

Huygen’s Principle Huygen first explained this in 1678 by proposing that all planar wavefronts are made up of lots of spherical wavefronts.. That is, you see how light propagates by breaking a wavefront into little bits, and then draw a spherical wave emanating outward from each little bit. You then can find the leading edge a little later simply by summing all these little “wavelets”

Huygen’s Principle Huygen first explained this in 1678 by proposing that all planar wavefronts are made up of lots of spherical wavefronts.. That is, you see how light propagates by breaking a wavefront into little bits, and then draw a spherical wave emanating outward from each little bit. You then can find the leading edge a little later simply by summing all these little “wavelets” It is possible to explain reflection and refraction this way too.

Diffraction at Edges what happens to the shape of the field at this point?

Diffraction at Edges I Light gets diffracted at the edge of an opaque barrier  there is light in the region obstructed by the barrier

As The Wave Propagates Out Spherically Its Intensity Decreases.....this happens with an edge too.. Diffraction places a finite limit on the formation of images

Double-Slit Interference Because they spread, these waves will eventually interfere with one another and produce interference fringes

Double-Slit Interference

Bright fringes

Double-Slit Interference Bright fringes screen

Double-Slit Interference Bright fringes Thomas Young (1802) used double-slit interference to prove the wave nature of light. screen

L d y r1r1 r2r2  P Double-Slit Interference Light from the two slits travels different distances to the screen. The difference r 1 - r 2 is very nearly d sin . When the path difference is a multiple of the wavelength these add constructively, and when it’s a half-multiple they cancel.

L d y r1r1 r2r2  d sin  = m  bright fringes d sin  = ( m+1/2) dark fringes P Double-Slit Interference Light from the two slits travels different distances to the screen. The difference r 1 - r 2 is very nearly d sin . When the path difference is a multiple of the wavelength these add constructively, and when it’s a half-multiple they cancel.

y d sin  = m  bright fringes d sin  = ( m+1/2) dark fringes Now use y = L tan  ; and for small y  sin   tan  = y / L y bright = m L/d y dark = (m+ 1/2) L/d L d r1r1 r2r2  P Double-Slit Interference Light from the two slits travels different distances to the screen. The difference r 1 - r 2 is very nearly d sin . When the path difference is a multiple of the wavelength these add constructively, and when it’s a half-multiple they cancel.

L d y r1r1 r2r2  P Intensity in Double Slit Interference Pattern This expression can be transformed using: sin  + sin  = 2 [sin (  +  ) / 2] [cos (  -  ) / 2] with  =  t and  =  t +  The electric field at P is E = E 1 + E 2 E = E P sin (  t) + E P sin (  t +  ) E = E P [sin (  t) + sin (  t +  )] E = 2 E p cos (  /2) sin (  t +  /2)

L d y r1r1 r2r2  P Intensity in Double Slit Interference Pattern The electric field at P is E = E 1 + E 2 E = 2 E p cos (  /2) sin (  t +  /2) Path difference is d sin    / 2  = d sin  /  E = 2 E P cos [  d sin  / ] sin (  t +  /2)

L d y r1r1 r2r2  P Intensity in Double Slit Interference Pattern The electric field at P is E = E 1 + E 2 E = 2 E P cos [  d sin  / ] sin (  t +  /2) E = E P * sin (  t +  /2) with E P * = 2 E P cos [  d sin  / ] The average intensity is derived from the Poynting vector S = (E x B) /  0 = E P B P /  0  (E P * ) 2 / 2  0 c

L d y r1r1 r2r2  P Intensity in Double Slit Interference Pattern The electric field at P is E = E 1 + E 2 E = E P * sin (  t +  /2) with E P * = 2 E P cos [  d sin  / ] The average intensity at P is = (E P * ) 2 / 2  0 c

L d y r1r1 r2r2  P Intensity in Double Slit Interference Pattern Phase is k(r 1 -r 2 ):

Example: Double Slit Interference 50 cm d Light of wavelength = 500 nm is incident on a double slit spaced by d = 50  m. What is the fringe spacing on the screen, 50 cm away?

Example: Double Slit Interference 50 cm d Light of wavelength = 500 nm is incident on a double slit spaced by d = 50  m. What is the fringe spacing on the screen, 50 cm away?

With more than two slits, things get a little more complicated L d y P Multiple Slit Interference

With more than two slits, things get a little more complicated L d y P Now to get a bright fringe, many paths must all be in phase. The brightest fringes become narrower but brighter; and extra lines show up between them.

Now to get a bright fringe, many paths must all be in phase. The brightest fringes become narrower but brighter; and extra lines show up between them. Multiple Slit Interference With more than two slits, things get a little more complicated L d y P Such an array of slits is called a “Diffraction Grating”

All of the lines show up at the set of angles given by: d sin  = (m/N) (N = number of slits). Most of these are not too bright. The very bright ones are for m a multiple of N. We won’t worry about the math here, just look at the general form: S  Multiple Slit Interference

Angular Spread:  ~  / a I Intensity distribution Single Slit Diffraction Each point in the slit acts as a source of spherical wavelets Slit width a  For a particular direction , wavelets will interfere, either constructively or destructively, resulting in the intensity distribution shown.

Single Slit Diffraction Each point in the slit acts as a source of spherical wavelets For a particular direction , wavelets will interfere, either constructively or destructively. Result: I This gives the angular spread ~ /a. This is for a slight of width a.

Example: Single Slit Diffraction a 50 cm Light of wavelength = 500 nm is incident on a slit a=50  m wide. How wide is the intensity distribution on the screen, 50 cm away?

Example: Single Slit Diffraction a 50 cm Light of wavelength = 500 nm is incident on a slit a=50  m wide. How wide is the intensity distribution on the screen, 50 cm away?

Example: Single Slit Diffraction a 50 cm Light of wavelength = 500 nm is incident on a slit a=50  m wide. How wide is the intensity distribution on the screen, 50 cm away? What happens if the slit width is doubled?

Example: Single Slit Diffraction a 50 cm Light of wavelength = 500 nm is incident on a slit a=50  m wide. How wide is the intensity distribution on the screen, 50 cm away? What happens if the slit width is doubled? The spread gets cut in half.

The Diffraction Limit Diffraction imposes a fundamental limit on the resolution of optical systems: S1S1 S2S2 Suppose we want to image 2 distant points, S 1 and S 2, through an aperture of width a: Two points are resolved when the maximum of one is at the minimum of the second The minima occurs for sin  = / a a = slit width Using sin      min = / a  L D D min / L  / a

The Diffraction Limit Diffraction therefore imposes a fundamental limit on the resolution of optical systems:  S1S1 S2S2 Suppose we want to image 2 distant points, S 1 and S 2 through an aperture of width a: D L The image separation is D ~ Lsin  L . The image blur is B ~ L /a To resolve individual points, we want: separation > blur so  > /a B