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Numerical-1

Numerical-1. In Lloyd’s single mirror interference experiment, the slit source is at a distance 2 mm from the plane of mirror. The screen is kept at a distance of 1.5 m from the source. Calculate the fringe width for the wavelength 6000 Å of light used. Ans: 0.225mm β = λ D/d.

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Numerical-1

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  1. Numerical-1 • In Lloyd’s single mirror interference experiment, the slit source is at a distance 2 mm from the plane of mirror. The screen is kept at a distance of 1.5 m from the source. Calculate the fringe width for the wavelength 6000 Å of light used. Ans: 0.225mm β=λD/d

  2. Phase change on Reflection, Refraction

  3. Principle of optical reversibility In the absence of any absorption, a light ray that is reflected or refracted will retrace its original path if its direction is reversed. a  amplitude of incident ray r1 reflection coefficient t1,  transmission coefficient ar1 amplitude of reflected ray at1 amplitude of refracted ray Note r1, r2 reflection coefficients t1, t2  transmission coefficients n1, n2  refractive index of two media

  4. According to principle of optical reversibility the two rays of amplitudes ar12 and at1t2 must combine to give the incident ray of fig 1. So, ar12 + at1t2 = a Stokes’ relation  t1t2 = 1- r12 ………(1) The two rays of amplitudes at1r1 and at1r2 must cancel each other. So, at1r1 + at1r2 = 0 Stokes’ relation  r2 = -r1 ……………(2)  t12+ r12 =1 ………..(3) ………..(4)  t2 = t1

  5.  t2 = t1  r2 = -r1 Stokes’ relation The negative sign in the amplitude of reflected wave indicates the phase difference of π b/w two rays The wave looses a half wave on reflection at the boundary of rarer to denser medium: π phase or λ/2 path difference Phase change  occurs when light gets reflected from denser medium.  LLOYD’S MIRROR no phase change will occur when light gets Reflected by a rarer medium.

  6. Interference in thin films due to reflection (Division of Amplitude) Colors of oil film on water Colors of soap bubble Antireflection thin films

  7. Interference byDivision of Amplitude Examples • Interference by thin films • Newton’s Ring • Michelson’s Interferometer

  8. For thin-film optics: • tthin-film1 µm • ( tthin-film order of the visible light)

  9. Condition for Constructive interference 2 tcos(r) - /2 = n 2 tcos(r) = (2n+1) /2 ……(1) Destructive interference 2 tcos(r)=n ……(2) Interference for Uniform thickness ‘t’ Show that Reflected Rays: Transmitted Rays:

  10. Interference by the film of Uniform thickness ‘d’: Reflected rays Path difference due to optical path b/w beam 1 & 2 Δ1 = Path ACD(in thin film) – path AB (in air) AB = AD sin (i) = 2AE sin(i) = 2d tan(r)*sin(i) = 2d tan(r)* μ*sin(r) =2 μ d sin2(r)/cos(r) Δ1 = μ(AC+AD) – AB = μ( d/cos(r)+ d/cos(r)) – AB Δ1 = 2μd/cos(r) – 2 μ d sin2(r)/cos(r) Δ1 = 2μ*d*cos(r)

  11. Interference by the film of Uniform thickness ‘d’: Reflected rays Path difference due to optical path b/w beam 1 & 2 Δ1 = 2μ*d*cos(r) Path difference due to the phase change in beam 1 along AB, after reflection from denser surface Δ2 = λ/2 Total path difference b/w 1 & 2 Δ = 2μ*d*cos(r) - λ/2

  12. Interference by the film of Uniform thickness ‘d’: Reflected rays Total path difference b/w 1 & 2 Δ = 2μ*d*cos(r) - λ/2 Condition for maxima Δ = 2μ*d*cos(r) - λ/2 = n λ 2 dcos(r) = (2n+1) /2 n = 0,1,2,3… Condition for minima Δ = 2μ*d*cos(r) - λ/2 = (2n-1) λ/2 2 dcos(r) = n  n = 0,1,2,3…

  13. Interference by the film of Uniform thickness ‘d’: Transmitted rays: Do it your self….. Condition for maxima 2 dcos(r) = n n = 0,1,2,3… Condition for minima 2 dcos(r) = (2n+1) /2 n = 0,1,2,3…

  14. Colors in thick films • A thick film shows no color in reflected system when illuminated with an extended source of white light. • An excessively thin film appears black in reflected system when illuminated with an extended source of white light

  15. Problem: A man whose eyes are 150 cm above the oil film on water surface observes the greenish color at a distance of 100 cm from his feet. Calculate the probable thickness of the film. ( =500nm, oil=1.4 , water=1.33) 9.725×10-6(2n-1) cm Problem: A parallel film of sodium light of wavelength 5880Å is incident on a thin glass plate of =1.5 such that the angle of refraction in the plate is 60o. Calculate the smallest thickness of the plate which will make it appear dark at reflection. 3920Å

  16. Numerical • Light of wavelength 589.3nm is reflected at nearly normal incidence from a soap film of refractive index 1.42. what is the least thickness of the film that will appear (I) black, (II) bright? • Ans: 207.5 nm, 103.75 nm. • A parallel beam of sodium light (589nm) is incidenting on an oil film on water. When viewed at an angle of 300 from normal, 8th dark band is seen. What is the thickness of oil film? • Ans: 1.7 m

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