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InterviewSolution
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Define Snell's Law. Using a neat labelled diagram derive an expression for the refractive index of the material of an equilateral prism. |
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Answer» Solution :Snell's law : The ratio of the sine of the ANGLE of incidence to the sine of angle of refraction is constant, called the refractive index of the medium. `(sini)/(sinr)=mu"(constant)".` Let ABC be the glass prism. Its angle of prism is A. The refractive index of the material of the prism is `mu`. Let AB and AC be the two refracting surfaces PQ = incident ray, RS = EMERGENT ray. `"Let angle of incidence "= r_(1)"angle of emergence "=i_(2)` `"angle of refraction "=r_(1)"angle of refraction at R"=r_(2)` After travelling through the prism it falls on AC and emerges as RS. The D = angle of deviation. From the `DeltaQRT` `r_(1)+r_(2)+angleT=180^(@)"..................(1)"` From the quadrilateral AQTR `angleA+angleT=180^(@)` `angleT=180^(@)-A."..............(2)"` From the equations (1) and (2) `r_(1)+r_(2)+angleT=180^(@)" we get "` `r_(1)+r_(2)+180^(@)-A=180^(@)` `r_(1)+r_(2)=A."...............(3)"` from the `DeltaQUR` `i_(1)-r_(1)+i_(2)-r_(2)+180^(@)-D=180^(@)` `i_(1)+i_(2)-(r_(1)+r_(2))=D` `i_(1)+i_(2)-A=D""[because r_(1)+r_(2)=A]` `i_(1)+i_(2)=A+D"....................(4)"`  MINIMUM deviation : Experimentally it is found that as the angle of incidence increased the angle of deviation decreases till it reaches a minimum value and then it increases. This least value of deviation is called angle of minimum deviation `'delta'` as shown in the fig. When D decreases the two angles `i_(1) and i_(2)` become closer to each other at the angle of minimum deviation, the two angles of incidence are same i.e., `i_(1)=i_(2)` As `i_(1)=i_(2), r_(1)=r_(2)` `therefore i_(1)=i_(2)=i, r_(1)=r_(2)=r,` substituting this in (1) and (2) we get `"2r = A"rArr r=A//2` `i+i=A+delta rArr i=(A+delta)/(2)` According to Snell's law `mu=("Sin i")/("Sin r")` `mu=("sin"((A+delta)/(2)))/("Sin A/2")` 
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