Interference

(1) Intensity of interference pattern,
$$I_{max} / I_{min} = (a+1)^2 / (a-1)^2 where, a>1$$
where, $$I_{max} = \text"Maximum Intensity" $$

  $$ I_{min} = \text"Minimum Intensity" $$

(2) Reflected light,

  (i) For maxima or bright fringe,

  $$ 2μtcosr = (2n-1) λ/2 $$

where, n = 1,2,3,…
  μ = Refractive index
  λ = Wavelength of light
  r = Angle of refraction

  (ii) For minima or dark fringe,

$$ 2μtcosr = nλ $$
   n = 0,1,2,3,..

(3) Fringe width,

$$ ω = λ/{2μθ} $$

where, λ = Wavelength of light
  θ = Angle of wedge
  ω = Width of fringes

(4) Snell’s law,

$$ μ = {sin i}/{sin r} $$

where, μ = Refractive index
  i = Angle of incidence

(5) Thickness of soap film,

$$ t = λ/{4μ} \text"(for maxima and normal incidence)" $$

where, λ = Wavelength of light
  t = Thickness of film
  μ = Refractive index

(6) In Newton’s ring experiment

  (6A) Diameter of dark ring,
$$ D_n = {4nRλ}/{μ} $$

where, $ D_n_^2 = \text"Diameter of " n^{th} \text" bright ring."$
  λ = Wavelength of light
  R = Radius of curvature of lens
  μ = Refractive index

  (6B) Diameter of bright ring is,

$$ D_n_^2 = {2(2n-1)λR}/{μ}$$
  $D_n = \text"Diameter of " n^{th} \text" bright ring."$
  λ = Wavelength of light
  R = Radius of curvature of lens
  μ = Refractive index

(7) For the measurement of wavelength and refractive index,
  (i) $$ λ = {μ(D_{n+p}_^2 – D_n_^2)}/{4pR} $$
  (ii) $$ μ = {4pRλ}/{D_{n+p}_^2 – D_n_^2} $$
    i.e. $ D_{n+p}_^2 – D_n_^2 = {4pRλ}/{μ} $