# HOW THE INTENSITY AND AMPLITUDE OF REFRACTED RAYS DIFFERS FROM THE INCIDENT RAY ??? PLZ EXPLN IN DETAIL.......

valentin68 | College Teacher | (Level 3) Associate Educator

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Basically when a ray of light strikes a surface that separates two medium, a part of the incoming ray is reflected back into the initial medium while another part of the light ray is transmitted. Because light can be regarded as an electromagnetic wave, at the separating surface continuity conditions for the E and B vectors of light need to be imposed to find the coefficients of reflection and transmission. For electric vector these continuity conditions on to the x (parallel to separation plane) and y axis (perpendicular to separation plane) writes as

`E_(x1) =E_(x2)`` `  (1)

`n_1*E_(y1) =n2*E_(y2)`  (2)

The first equation means the parallel component of E is the same in the separation plane (otherwise hypothetical electric charges will be moved in the separation plane) and the second equation means the fact that there is no electric charge present in the separation plane (otherwise the term `Q/epsilon` need to be added to the right term of the expression- see Gauss law). Here `n_1` and `n_2` are the refraction indexes of the two different mediums.

The above equations means the reflection coefficient (and also the transmission coefficient) will depend on the type of polarisation the incoming light ray has. Usually these coefficients are given for the case when the vector `E` is either perpendicular or parallel to the plane that contains both reflected and refracted waves. Their values are (see Fresnel equations)

`T_("parallel") =(sin(2alpha_i)*sin(2alpha_r))/(sin^2(alpha_i+alpha_r)*sin^2(alpha_i-alpha_r))`

`T_("perpendicular") =(sin(2*alpha_i)*sin(2alpha_r))/(sin^2(2*alpha_i))`

where `alpha_i and alpha_r` are the incidence and refraction angles.

Now the answer to your question is simple. The rapport of transmitted and incident intensities (or of the squares of amplitudes) is equal to `T` coefficient.

`T =I_r/I_i = (E_r/E_i)^2`

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