Determine the reflectance from tin at angles of incidence of 0^∘, 30^∘, and 60^∘ . Do this for t

Step 1: Calculate the angle of refraction using the formula $sin theta_2 = frac{n_1}{n_2} si

Determine the reflectance from tin at angles of incidence of...

Determine the reflectance from tin at angles of incidence of $0^{\circ}, 30^{\circ},$ and $60^{\circ} .$ Do this for the (a) TE and (b) TM modes of polarization. Real and imaginary parts of the complex refractive index are 1.5 and $5.3,$ respectively, for light of $589.3 \mathrm{nm}$

Transcript

00:01 Hi let's determine the reflectance from tin at angle of incident of 0 degree 30 degree and 60 degree to this for the te and tm modes of polarization real and imaginary part of the complex reflective index are 1 .5 and 5 .3 respectively for light of 589 .3 nanometer the formula for reflectance for tm incident tm is stands for transverse magnetic mode which is n2 cos theta 1 minus n1 cost teta 2 divided by n2 cos theta 1 plus n1 cos theta 2 similarly for t e incident r e equal to n1 cos theta 1 plus j n2 root of n1 by n2 the whole square into sine square theta 1 minus 1 divided by n1 cross theta 1 minus j n2 root over n 1 by n2 the whole square sine square teta 1 minus 1 here are this reflectance and n1 is refractive index of first medium and n2 is refractive index of second medium and teta 1 is an angle of in theta 2 is an angle of refraction to calculate reflectance first calculate theta 2 using following formula which is n1 sine theta 1 equal to n2 sine teta 2 so teta 2 equal to sine inverse n 1 by n 2 sign theta 1 now using this formula we're going to calculate what is theta 2 now i'm going to substitute n1 equal to 1 .5 and n2 equal to 5 .3 then teta equal to 0 degree so we have theta 2 equal to sine inverse of 1 .5 by 5 .3 sign 0 degree which is equal to 0 degree.

03:04 Now substitute n1 equal to 1 .5 n2 equal to 5 .3 and teta equal to 30.

03:14 The same thing i'm going to substitute all the value instead of theta already i have calculated 0 degree now i'm going to do for 30 degrees so i will be getting theta 2 equal to 8 .14 degree now i'm going to substitute theta equal to 60 degree then i will be getting theta 2 equal to 14 .19 degree now calculate rm for theta 1 equal to 0 degree here i'm going to substitute n1 equal to 1 .5 n2 equal to 5 .3 teta 1 equal to 0 degree and teta 2 equal to 0 degree so rm equal to 5 .3 into 1 minus 1 .5 into 1 divided by 5 .3 into 1 plus 1 .5 into 1 divided by 5 .3 into 1 plus 1 .5 into 1 which is equal to 0 .56 similarly are e equal to 1 .5 into 1 .5 into 1 plus j 5 .3 into root of 1 .5 by 5 .3 is the whole square into 0 square minus 1 divided by 1 .5 into 1 minus j into 5 .3 root of 1 .5 by 5 .5 by 5 .3 the whole square into 0 square minus 1 which is equal to 1 .5 minus 5 .3 divided by 1 5 .5 plus 5 .3 equal to minus 0 .56...

Question

Determine the reflectance from tin at angles of incidence of $0^{\circ}, 30^{\circ},$ and $60^{\circ} .$ Do this for the (a) TE and (b) TM modes of polarization. Real and imaginary parts of the complex refractive index are 1.5 and $5.3,$ respectively, for light of $589.3 \mathrm{nm}$

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