- #1
Anabelle37
- 35
- 0
URGENT! michelson interferometer - maxima and minima
If a wave is reflected at a surface of a plate with higher refractive index (eg. At air glass or air-metal) it suffers a λ/2 phase change. Show that for the interferometer, maxima will occur for 2dcosθ = (n+1/2)λ and minima for 2dcosθ = nλ.
When the path length difference is zero or an integer multiple of the wavelength, the waves arrive at a common point exactly in phase and they interfere fully constructively there. This is seen by the bright fringes, so maxima occur when 2dcosθ = nλ (n=0,1,2...) (1)
But because the wave is reflected at a surface of a plate with higher refractive index, it suffers a λ/2 phase change. So equation (1) becomes 2dcosθ = nλ + λ/2 = (n+1/2)λ (2)
However, when the path length difference is an odd multiple of half the wavelength, the waves arrive at a common point exactly out of phase and they interfere fully destructively there. This is seen by the dark fringes, so minima occur when 2dcosθ = (n+1/2)λ (3)
But because the wave is reflected at a surface of a plate with higher refractive index, it suffers a λ/2 phase change. So equation (3) becomes 2dcosθ = (n+1)λ (4)
Equation 4 is not correct so I don't know if I'm going about it the right way??
PLEASE HELP!
Homework Statement
If a wave is reflected at a surface of a plate with higher refractive index (eg. At air glass or air-metal) it suffers a λ/2 phase change. Show that for the interferometer, maxima will occur for 2dcosθ = (n+1/2)λ and minima for 2dcosθ = nλ.
Homework Equations
The Attempt at a Solution
When the path length difference is zero or an integer multiple of the wavelength, the waves arrive at a common point exactly in phase and they interfere fully constructively there. This is seen by the bright fringes, so maxima occur when 2dcosθ = nλ (n=0,1,2...) (1)
But because the wave is reflected at a surface of a plate with higher refractive index, it suffers a λ/2 phase change. So equation (1) becomes 2dcosθ = nλ + λ/2 = (n+1/2)λ (2)
However, when the path length difference is an odd multiple of half the wavelength, the waves arrive at a common point exactly out of phase and they interfere fully destructively there. This is seen by the dark fringes, so minima occur when 2dcosθ = (n+1/2)λ (3)
But because the wave is reflected at a surface of a plate with higher refractive index, it suffers a λ/2 phase change. So equation (3) becomes 2dcosθ = (n+1)λ (4)
Equation 4 is not correct so I don't know if I'm going about it the right way??
PLEASE HELP!