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x-y plane separates two media, zge0contains a medium of refractive index 1 and zle0contains a medium of refractive index 2. A ray of light is incident from first medium along a vectorhati+hatj-hatk. Find the unit vector along the refracted ray. |
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Answer» Solution :The law of refraction in vector form is `mu_(1)(hat(e)_(1)xxhat(N))=mu_(2)(hat(e)_(2)xxhat(n))` Here `hat(e)=(hat(i)+hat(J)-hat(k))/(sqrt(3))` `=` unit vectore along incident ray `hat(n)=hat(k)=` unit vector along normal or incidence point `:. mu_(1)(hat(e)_(1)xxhat(n))=mu_(2)(hat(e)xxhat(n))` `hat(e)_(2)=xhat(i)+yhat(j)+zhat(k)` `=` unit vectore along REFRACTED ray or `1{((hat(i)+hat(j)-hat(k))/(sqrt(3)))xxhat(k)}=2{(xhat(i)+yhat(j)+zhat(k))xxhat(k)}` or`(-hat(j)+hat(i))/(sqrt(3))=2{-xhat(I)+yhat(j)}` or `x= (1)/(2sqrt(3)), y=(1)/(2sqrt(3))` As `hat(e)_(2)` is a unit vector, THEREFORE `|hat(e)_(2)|=1` `RARR sqrt(x^(2)+y^(2)+z^(2))=1` or `sqrt((1/(2sqrt(3)))^(2)+((1)/(2sqrt(3)))+z^(2))=1` or ` sqrt(((1)/(12)+(1)/(12)+z^(2)))=1` or `(1)/(6)+z^(2)=1` `:. z^(2)= 1-(1)/(6)=(5)/(6)` `:. z=+-sqrt(((5)/(6)))` Since refracted ray is in negative z-axis region `:. z=-sqrt((5)/(6))` `:. hat(e)_(2)=(1)/(2sqrt(3))hat(i)+(1)/(2sqrt(3))hat(j)-sqrt(((5)/(6)))hat(k)` 
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