Question

What is the phase shift of the TM and TE rays reflected both externally (air to glass) and internally (glass to air) for the situation discussed in Prob. 10?
Prob. 10 is Calculate the reflectance and transmittance for both TE and TM modes of light incident from air at 30 onto glass of index 1.60.

Answer 1

Given,

Angle of incidence, _i = 30

Refractive index of air, n₁ = 1

Refractive index of glass, n₂ = 1.60

Let angle of transmission be _t

Now,

as we know that, according to Snell's law

n₁*sin_i = n₂*sin_t

=> 1 * sin30 = 1.6 * sin_t

=> sin_t = 0.5 / 1.6 =0.3125

=>    _t = sin^-1 (0.3125) = 18.21

or    _t = 18.21

Now,

Reflectance for TE mode is given by

r_TE = (n₁ cos_i - n₂ cos_t ) / (n₁ cos_i + n₂ cos_t )

=>        = (n₁ cos30 - n₂ cos(18.21) ) / (n₁ cos30 + n₂ cos18.21 )

=>        = ( 0.866 - (1.60 * 0.95)) / ( 0.866 + (1.60 * 0.95))

=>        = - 0.654 / 2.386 = -0.274

Since, r_TE is negative, hence _TE (_i = 30) is 180

Now,

Reflectance for TM mode is given by

r_TM = (n₂ cos_i - n₁ cos_t ) / (n₂ cos_i + n₁ cos_t )

=>   = (n₂ cos30 - n₁ cos(18.21) ) / (n₂ cos30 + n₁ cos18.21 )

=>   = ((1.6 * 0.866) - 0.95 ) / ((1.6 * 0.866) + 0.95 )

=>    = 0.4356 / 2.3356 = 0.187

Since, r_TM is positive , hence _TM (_i = 30) is 0

Now,

Transmittance for TE mode is given by

t_TE = 2n₁ cos_i / ( n₁ cos_i + n₂ cos_t )

=>   = 2 * 1 *cos30 / ( 1*cos30 + 1.6 * cos(18.21))

=>    = 2*0.866 / ( 0.866 + 1.60 * 0.95)

=>    = 1.732 / 2.386 = 0.726

Since, t_TE is positive, hence _TE (_i = 30) is 0

Now,

Transmittance for TM mode is given by

t_TM = 2n₁ cos_i / ( n₂ cos_i + n₁ cos_t )

=>   = 2 * 1 *cos30 / ( 1.60*cos30 + cos(18.21))

=>   = 1.732 / ( 1.60 * 0.866 + 0.95 )

=>   = 1.732 / 2.3356 = 0.742

Since, t_TM is positive, hence _TM (_i = 30) is 0

Now,

For glass to air

n₁ = 1.6

and n₂ = 1

and _i = 18.21 and _t = 30

hence,

r_TE = (n₁ cos_i - n₂ cos_t ) / (n₁ cos_i + n₂ cos_t )

=>        = (1.6*cos18.21 - 1*cos30 ) / (1.6*cos18.21 + 1*cos30 )

=>        = 0.654 / 2.386 = 0.274

Since, r_TE is positive, hence _TE (_i = 18.21) is 0

Now,

r_TM = (n₂ cos_i - n₁ cos_t ) / (n₂ cos_i + n₁ cos_t )

      = (1*cos18.21 - 1.6*cos30 ) / (1*cos18.21 + 1.6*cos30 )

      = -0.4356 / 2.3356 = -0.187

Since, r_TM is negative, hence _TM (_i = 18.21) is 180

Now,

Transmittance for TE mode is given by

t_TE = 2n₁ cos_i / ( n₁ cos_i + n₂ cos_t )

=> = 2 * 1.6 *cos18.21 / ( 1.6*cos18.21 + 1*cos30 )

=>   = 3.04 / 2.386 = 1.274

Since, t_TE is positive, hence _TE (_i = 18.21) is 0

Now,

Transmittance for TM mode is given by

t_TM = 2n₁ cos_i / ( n₂ cos_i + n₁ cos_t )

=>   = 2 * 1.6 *cos18.21 / ( 1*cos18.21 + 1.60*cos30)

=>   = 3.04 / 2.3356 = 1.302

Since, t_TM is positive, hence _TM (_i = 18.21) is 0