A ray is incident on one face of a triangular glass prism inair. The angle of in

Question

A ray is incident on one face of a triangular glass prism inair. The angle of incident is chosen so that theemerging ray also makes the same angle with the normal tothe other face. Show that the index of refraction n of the glass prism is given by: where is the vertex angle of the prism and is the deviation angle, the total angle through which the beam isturned in passing through the prism. (Under these conditions thedeviation angle has the smallest possible value, which iscalled the angle of minimum deviation)

Explanation / Answer

   for simplicity we note that the light rays pointof entry A, the vertex of prism B, exit point is C    in the given figure in the text book where is defined is denoted as D    the angle indicated by ADC is the supplementof so we denote that    s = 180o -    the angle of refraction in the glassis    2 = (1 / n) sin    the angle between trhe interior ray and thenear by surfaces is the complement of 2 so wedenote it as    2c = 90o -2    so that the angles in the trianglemust addto 180o    180o + 2 2c +    2 = / 2    the angles in the triangle ADC must add to180o so that    180o = 2 ( -2) + s     = 90o + 2- (1 / 2) s      = 2 + (1 / 2) s       = (1 / 2) ( +)    so the index of refraction will be    n = sin / sin2       = sin((1 / 2) ( +)) / sin((1 / 2) )      = 2 + (1 / 2) s       = (1 / 2) ( +)    so the index of refraction will be    n = sin / sin2       = sin((1 / 2) ( +)) / sin((1 / 2) )

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