InterviewSolution
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InterviewSolution
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Show thatn=(sin((A+D)/(2)))/(sin((A)/(2))) where symbols have their usual notations. |
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Answer» Solution :BC - Base of the prism AB and AC - Refracting sides representing the planes ABC - Principle section of prism. PQ - incident ray. RS - emergent ray `d_(1)` - angle of deviation `d_(1)=i_(1)-r_(1)` `d_(2) ` - angle of deviation`d_(2)=i_(2)-r_(2)` `u hatTS `- d - total angle of deviation or net deviation i.e., `d=d_(1)+d_(2)` (exterior angle = sum of two opposite interior angles)  NM is normal to AB at Q N.M is normal to AC at R `i_(1)`- angle of incidence `i_(2)` - angle of emergence. D = angle of minimum deviation, From figure (1) From the quadrilateral AQMR, `A hatQM~=A hatRM=90^(@)` (Nm, N.M are normal to AB and BC) So that `A+M=180^(@)""` ......(1) `THEREFORE `AQMR is a cyclic quadrilateral. From the triangle QMR `r_(1)+r_(2)+hatM=180^(@)("Property of a "Delta) "" `......(2) From (1) and (2) `A+M=r_(1)+r_(2)+M` i.e.,`A=r_(1)+r_(2)""`.......(3) For non-symmetric condition `i_(1) ne i` `d_(1) ne d_(2)` also net deviationd. `d=d_(1)+d_(2)` `d=i_(1)-r_(1)+i_(2)-r_(2)` `therefored=i_(1)+I_(2)-(r_(1)+r_(2))` using (3) we write, `d=i_(1)+i_(2)-A""`.......(4) from the fig., we note that for non-symmetric condition for two angles of incidence, angle of net deviation is the same. However for a symmetric condition, `i_(1)=i_(2)=i and r_(1)=r_(2)=r` net deviation becomes the minimum angle of deviation. In this case the refracted ray will be parallel to the base. `therefore` for symmetric condition, (4)can be written as `D=2i-A` i.e., `i=((A+D)/(2))""` ......(5) (3) can be written as, `A=2r` or`r=((A)/(2))"" ` ......(6) From Snell.s law of refraction at AB, |
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