Unit 1 Interference PDF

## Coherent Beams A steady interference pattern can be obtained if the two light waves superimposing on each other are coherent. The phase difference between them is always constant. ### Methods of getting coherent Beams 1. **Division of wavefront method** - Incident wavefront is divided into tw...

## Coherent Beams A steady interference pattern can be obtained if the two light waves superimposing on each other are coherent. The phase difference between them is always constant. ### Methods of getting coherent Beams 1. **Division of wavefront method** - Incident wavefront is divided into two separate wavefronts which then superimpose to give the steady interference pattern. - Ex: Young's Experiment. 2. **Division of amplitude method** - Incident wave is divided into two beams by using phenomenon of partial reflection and refraction. - Ex: Interference in thin film. ### Useful Relations 1. **Relation between path difference and phase difference:** - Relation between path difference and phase difference for any two waves is given by: Φ = $2\pi$ (path diff) - Where Φ = phase diff. 2. **When ray of light reflected at surface of an optically denser medium while it is travelling through rarer medium,** - Phase change of $\pi$ or path diff of $\pi/2$ introduced. 3. **When light wave traverses a distance t in medium of R.I u, then equivalent optical path is given by ut.** ## Interference in Thin Films ### Interference Due to Reflected Light - Path difference - Assume ray of monochromatic light SA of wavelength λ from an extended source be incident on upper part surface of thin transparent film of thickness t and R.I. μ at an angle i'. - Ray SA is partly reflected along AB and partly refracted along AC at angle r. - Refracted part AC reflected from lower surface of the film along CD. - And finally emerges out along DE. - Ray AB and DE are derived from same incident ray so they are coherent. - To calculate path difference between AB and DE, perpendicular DL and CN are drawn on AB and AD respectively. - Path difference between AB and DE given by: Δ = Path (AC + CD) in film - Path (AL) in air Δ = μ (AC + CD) - AL - In right angled triangle ACN, cosγ = CN / AC AC = CN / cosγ - Where, LACN = γ and CN=t, thickness of film - Similarly, in other right angled ACMD, cosγ = CN / CD CD = CN / cosγ - In right angled AALD cosr = AL / AD sinr = AL / AD - Again in ΔACN & ΔCND tanr = AN / NC tanr = ND / NC - AN = NC tanr = t tanr - ND = NC tanr = t tanr - Put values of AN and ND in eqn(4), AL = (t tanr + t tanr) sinr AL = 2t tanr sinr AL = 2t sinr (sinr / cosr) AL = 2tμ sin²r / cosr - Substitute the values of AC, CD & AL from eq(2), (3) & (6) in eq(1) we get: Δ = μ [t/ cosr + t/ cosr] - 2ut sin²r / cosr Δ = 2ut / cosr - 2ut sin²r / cosr Δ = 2ut (1 - sin²r) / cosr Δ = 2ut cos²r / cosr Δ = 2ut cosr - As ray AB is reflected from denser medium so there occurs an additional path diff of π/2 or phase change π. - Total path diff. = 2ut cosr ± π/2 ### Condition for Constructive Interference or Maxima - In this case path diff, should be even multiple of π/2 - For minima: 2ut cosr + π/2 = 2n π/2 = nλ 2ut cosr = nλ - π/2 2ut cosr = (2n - 1) π/2 - Where n = 1, 2, 3 - 2ut cosr = nλ + π/2 2ut cosr = (2n + 1) π/2 - Where, n = 0, 1, 2,... ### Condition for Destructive Interference or Minima - In this case path diff should be an odd multiple of π/2. - For minima: 2ut cosr + π/2 = (2n ± 1) π/2 2ut cosr = πλ - π/2 2ut cosr = nλ - Where, n = 0, 1, 2,... - Therefore the conditions of maxima & minima in transmitted light are just opposite to those for reflected light. - Hence interference pattern of reflected and transmitted monochromatic light are Complementary. ## Colours in Thin Films 1. If an oil film on water or a soap film or wedge shaped air film between two glass plates is seen in reflected light, it shows brilliant colours. 2. This explanation of the origin of these coloured phenomenon is given by Young on basis of Interference of light waves. 3. If beam of light from extended source is incident normally on a thin film of transparent material and image observed reflected, then colour fringes will be seen. 4. Path diff. between these interfering rays depends upon thickness of the film and or incident ray or inclination of incident ray. 5. Condition for maxima is: 2ut cosr = (2n ± 1) π/2 6. White light consists of continuous range of wavelengths i.e. colours. 7. So only those colours which satisfy condition of maxima will be visible with maximum intensity. 8. Condition for minima is: 2ut cosr = nλ ## Interference in Wedge-shaped Film - Consider wedge-shaped film of R.I. μ enclosed by two plane surfaces OP and OQ inclined at angle θ. - The thickness of film increases from O to P. - Consider ray of light AB incident on film. - It will be partially reflected along BE and partially refracted along BC. - Ray BC will be partially reflected along CD and will emerge out along DF. - It is possible to see interference pattern through a microscope if angle of wedge θ is small. - The path difference between rays BE and DF is: Δ = path (BC+CD) in film - path (BG) in air Δ = μ (BC+CD) - BG - Where, μ = R.I of film. - By solving, it can be shown that this path diff is: Δ = 2ut cos(r + θ) - Ray DF suffers phase change of π due to reflection from denser medium at C. - Whereas there is no normal phase change due to reflection for rays BE, Δ = 2ut cos (r + θ) ± π/2 ### Condition for Constructive Interference or Maxima: - Δ = nλ - 2ut cos(r + θ) + π/2 = nλ - 2ut cos(r + θ) = nλ ± π/2 - 2ut cos (r + θ) = (2n + 1) π/2 ### Condition for Destructive Interference or Minima: - Δ = (2n + 1) π/2 - 2ut cos(r + θ) + π/2 = (2n ± 1) π/2 - 2ut cos (r + θ) = nλ ### Fringe Width - The distance between two successive bright fringes or two successive dark fringes is known as fringe width. - If nth bright/dark fringe is formed at distance xn from edge of wedge shape film where thickness of film t = tn. - From fig: tan θ = tn / dn tn = xn tan θ - Similarly, (n + 1)th dark fringe is formed at distance xn+1. tan θ = xn+1 / dn+1 xn+1 = dn+1 tan θ - We know the condition for destructive interference i.e. minima: 2ut cos(r + θ) = nλ - For normal incidence i = 0 ⇒ r = 0 - Above eqn will become: 2ut cos0 = nλ - Let consider above condition for nth dark fringe. - 2ut cos0 = nλ - From (2) tn = xn tanθ - 2ut xn tanθ cosθ = nλ - 2ut xn sinθ cosθ = nλ - For (n + 1)th dark fringe eq(3) become: 2ut xn+1 cosθ = (n + 1) λ - But from (2) xn+1 = xn+1 tan θ - 2ut xn+1 tanθ cosθ = (n + 1) λ - 2ut xn+1 sinθ cosθ = (n + 1) λ - 2ut xn+1 sin θ = (n + 1) λ - Subtracting eq (4) from eq (5) 2ut sin θ [xn+1 - xn] = (n + 1) λ - nλ 2ut sin θ [xn+1 - xn] = λ - But fringe width W = xn+1 - xn - 2ut sin θ W = λ - For very small angle θ: sin θ = θ - W = λ / 2utθ - For air film, μ = 1 - W = λ / 2θ - As we see in above eq, fringe width is independent of order n. - For all, fringes width will be equal. - The equation for fringe width can also be derived by using bright fringes. ## Anti- Reflection Coating - Anti- reflection (AR) coating is a type of optical coating applied to the surface of lenses, other optical elements and photovoltaic cells to reduce reflection. - In order to minimize loss of light, an antireflection coating is used. - The anti-reflection coatings are used in good cameras, projectors, binoculars and telescopes to prevent unwanted reflections from the surface. - The image formed by lens is less bright if loss of light due to reflection is more. - The surface of the glass is oftenly coated with a material is known as Magnesium Fluoride (MgF2), which is transparent and has R.I. between those of air and glass i.e. it has R1 = 1.38 - The thickness t' of film is chosen in such way that it provides destructive interference between the beam reflected from the surface of coating (ray 1) and that reflected from surface of lens (ray 2) as shown in fig. - If thickness of film is π/4 then two interfering ray 1 and 2 differ. - In path by π/2 (2 * π/4 * π/2), then it produces destructive interference for given wavelength. - The value of wavelength chosen around 5500 Å because eye is most sensitive to this wavelength. - For destructive interference: 2ut cosr = (2n - 1) π/2 - For minimum thickness, h = 1 2 μt cos γ = π/2 - For normal incidence, i = r = 0 2 μt = π/2 t = π/4μ