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# if a light ray is incident of the glass surface with index of refraction n=1.50 at 0 degree anglee.What would be the angle of refraction if

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if a light ray is incident of the glass surface with index of refraction n=1.50 at 0 degree anglee.What would be the angle of refraction if the index of refraction of air is n=1.00

Jul 23, 2014

#2
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Snell's law says that:  sin(θ1)/sin(θ2) = n2/n1 where the n's are the indices of refraction.  If the angle of incidence is zero, then the angle of refraction will be zero also, as the angles refer to the deviation from the normal to the surface.  However, I guess in your case you mean the angle from the surface, so that you are talking about grazing incidence.  In this case θ1 is actually 90° and sin(θ2) =  sin(90°)*1/1.5 = 1/1.5 so

$${\mathtt{theta2}} = \underset{\,\,\,\,^{{360^\circ}}}{{sin}}^{\!\!\mathtt{-1}}{\left({\frac{{\mathtt{1}}}{{\mathtt{1.5}}}}\right)} \Rightarrow {\mathtt{theta2}} = {\mathtt{41.810\: \!314\: \!895\: \!779^{\circ}}}$$

or θ2 ≈ 41.81° (Note that this is also the angle made with the normal to the surface.  The angle made with the surface will be 90° - 41.81°.)

Jul 23, 2014

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CAN ANYONE WRITE THE SOLUTION . THANKS

Jul 23, 2014
#2
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$${\mathtt{theta2}} = \underset{\,\,\,\,^{{360^\circ}}}{{sin}}^{\!\!\mathtt{-1}}{\left({\frac{{\mathtt{1}}}{{\mathtt{1.5}}}}\right)} \Rightarrow {\mathtt{theta2}} = {\mathtt{41.810\: \!314\: \!895\: \!779^{\circ}}}$$