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The figure shows a constant deviation prism ABCD. The incident ray is PQ and the emergent ray is ST. Although it is made up of one piece of glass, but it is equivalent to two 30^(@)-60^(@)-90^(@) prism and one 45^(@)-45^(@)90^(@) prism.The angle theta_(1) is the angle of incidence on face AB. The path of the ray inside the prism is indicated in the figure. for this prism mu=2sintheta_(1). Q. The total deviation of incident ray when it emerges out of the prism is |
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Answer» `60^(@)` `mu=(sintheta_(1))/(sinr),sintheta_(1)=(mu)/(2)impliessinr=(1)/(2),r=30^(@)` considering ABC as a triangular prism `r+r^(')=A` `30^(@)+r^(')=75^(@)impliesr^(')=45^(@)` THUS ray is incident on AB at `45^(@)` it is reflected at the same angle on to CD. From the figure `angleRSD=360^(@)-(45^(@)+135^(@)+60^(@))=120^(@)` (considering the quadrilateral) `because` angle of incidence at `CD=120^(@)-90^(@)=30^(@)` `impliestheta_(1)=theta_(2)` `because` total deviation of ray `=(theta_(1)-30^(@))+(180^(@)-90^(@))-(theta_(2)-30^(@))=90^(@)` |
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