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Let the x-y plane to be boundary between two transparent media. Medium 1 in Z ge0 has a refractive index of sqrt(2) and medium 2 with z lt 0 has a refractive of sqrt(3). A ray of light in medium 1 given by the vector vecA=6sqrt(3)hati+8sqrt(3)hatj-10hatk is incident onthe plane of separation. The angle of refracted ray in medium 2 is |
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Answer» SOLUTION :Unit vector representing the normal to the plane `hate_a=hatk`. Component of the INCIDENT ray along the normal is `-10 hatk` The unit vector that represents the plane of the incident ray and the normal. ` hate_p= ((6 sqrt(3) HATI + 8 sqrt(3)hatj))/( sqrt((6 sqrt(3))^2 +(8 sqrt(3))^2))=0.6hati + 0.8hatj` ANGLE between the incident ray and the normal is given by ` cos theta= ( 6 sqrt(3) hati + 8 sqrt(3) hatj - 10 hatk ) . hatk // sqrt((6 sqrt(3))^2 +(8 sqrt(3))^2 + 10^2)` (or)` cos theta=- 0.5 ` thereforetheangle` theta = 120^@` The angleofincidenceis ` theta= 180^@ - 120^@=60^@` theangleof therefractedbeamis givenby ` sqrt(2) sin ( theta) = sqrt(3) sin (r)orr= 45^@ ` The equationof theemergentrayis ` cos(45)(- hatk ) + sin(45)[0.6hati+ 0.8hatj]` `= 1// sqrt(2)( 0.6hati+ 0.8hatj - hatk)` |
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