cosine rule is 2μt cosr = nλ
it is the condition for dark fringe in interference due to reflected light from parallel surfaces.
μ= refractive index of the surface
r = angle of reflection , n = order of the fringe
and t = thickness of parallel film
λ = wavelength of monochromatic light.
The interference pattern consists a dark fringe at the center and surrounded by alternatively bright and dark fringes.
consider a parallel film of thickness t and refractive index μ.
Monochromatic light incidents on upper surface of film at an angle i.
A part of light is reflected along the path AB(light 1 ) from upper surface and the remaining light is refracted along the path AC through the film.
angle of refraction = r
the light AC is again reflected upwards from lower surface of the film in the direction CD and then emerges out from upper surface in the direction 2 ( light 2 ).
There is a path difference between the two lights 1 and 2.
BD is the normal line drawn on AB.
the light 1 travels an extra path in air which is equal to AB *1
refractive index of air =1
The second light 2 travels an extra path inside the film
equal to (AC + CD) * μ
hence the total path difference between the two lights is
Δ = ( AC + CD ) * μ - AB
it is clear that AC= CD and AC = EC / cosr
AC= t / cosr eq.1
(AC + CD )μ = 2t *μ / cosr
also AB = sin i * AD
where AD = AE + ED = t * tanr + t* tanr
AD = 2t (tanr) * sini
therefore path difference Δ = ( AC + CD ) * μ - AB
Δ = 2t *μ / cosr - 2t (tanr) * sini
applying snell's rule μ = sini/ sinr
sini =μ sinr
hence Δ = 2t *μ / cosr - 2t (tanr) * μsinr
Δ = 2t *μ / cosr - 2t (sinr/cosr) * μsinr
Δ = 2t *μ / cosr [ 1 - sin^2 r ]
Δ = 2t *μ / cosr [cos^2 r ]
Δ = 2μt cosr
it is the condition for dark fringe in interference due to reflected light from parallel surfaces.
μ= refractive index of the surface
r = angle of reflection , n = order of the fringe
and t = thickness of parallel film
λ = wavelength of monochromatic light.
The interference pattern consists a dark fringe at the center and surrounded by alternatively bright and dark fringes.
Monochromatic light incidents on upper surface of film at an angle i.
A part of light is reflected along the path AB(light 1 ) from upper surface and the remaining light is refracted along the path AC through the film.
angle of refraction = r
the light AC is again reflected upwards from lower surface of the film in the direction CD and then emerges out from upper surface in the direction 2 ( light 2 ).
There is a path difference between the two lights 1 and 2.
BD is the normal line drawn on AB.
the light 1 travels an extra path in air which is equal to AB *1
refractive index of air =1
The second light 2 travels an extra path inside the film
equal to (AC + CD) * μ
hence the total path difference between the two lights is
Δ = ( AC + CD ) * μ - AB
it is clear that AC= CD and AC = EC / cosr
AC= t / cosr eq.1
(AC + CD )μ = 2t *μ / cosr
also AB = sin i * AD
where AD = AE + ED = t * tanr + t* tanr
AD = 2t (tanr) * sini
therefore path difference Δ = ( AC + CD ) * μ - AB
Δ = 2t *μ / cosr - 2t (tanr) * sini
applying snell's rule μ = sini/ sinr
sini =μ sinr
hence Δ = 2t *μ / cosr - 2t (tanr) * μsinr
Δ = 2t *μ / cosr - 2t (sinr/cosr) * μsinr
Δ = 2t *μ / cosr [ 1 - sin^2 r ]
Δ = 2t *μ / cosr [cos^2 r ]
Δ = 2μt cosr
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