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Show that for a material with refractive index mu ge sqrt(2)~ , light incident at any angle shall be guided along a length perpendicular to the incident face. |
Answer» Solution : `rArr` As shown in the figure , consider a light ray `vec (PQ)`, MADE incident on the surface AB of a denser transparent medium, with angle of incidence i, After refraction, ray `vec(QR)` is incident at point R on the surface of RARER medium with angle of incidence `pi` . If `pi GE` C then light ray will have total inernal reflection again and again but it will not COM out of walls AC and BD. (Where C= cirtical angle of GIVEN denser medium w.r.t rarer medium).Thus, `phit ge C` `therefore90^@ - r ge C""(because phi + r + 90^@)` `therefore sin (90^@ - r) ge sin C` `cos r ge (1)/(mu) "" ( because sin C = 1/mu)` `thereforecos r ge 1`........(1) `rArr` Applying Snell.s law at point Q, (1) sin i `= mu sin r` `therefore sin r = (sin i)/(mu)`.......(2) `rArr` Now ` cos r sqrt( 1 - sin^2 r)` `= sqrt(1 - (sin^2 i)/(mu^2))` ` = sqrt(mu^2 - sin^2 i)/(mu)` `therefore mu cos r = sqrt(mu^2 - sin^2 i)/(mu)` `therefore cos r = sqrt(mu^2 -sin^2 i).........(3)` `rArr` From equation (2) and (3), `sqrt(mu^2 - sin^2i) ge 1` `thereforemu^2 - sin^2 i ge 1` `rArr` But maximum value of sin i is 1. Hence if above codition is satisfied for ` i = 90^@` then for all the value of i , above condition will be satisfied, Hence taking sin `i = sin 90^@ = 1,` `mu^2 - 1 ge 1` `therefore mu^2 ge 2` `therefore mu ge sqrt(2)` `rArr` Above relation gives required condition. |
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