Question

In above figure (a), a beam of light in material 1 is incident on a boundary at an angle of $$\theta _{1}=30^{0}$$. The extent of refraction of the light into material 2 depends, in part, on the index of refraction $$n_{2}$$ of material 2. Above figure (b) gives the angle of refraction $$\theta _{2}$$ versus $$n_{2}$$ for a range of possible $$n_{2}$$ values. The vertical axis scale is set by $$\theta _{2a}=20^{0}$$ and $$\theta _{2b}=40^{0}$$. (a) What is the index of refraction of material 1? (b) If the incident angle is changed to $$60^{0}$$ and material 2 has $$n_{2}=2.4$$, then what is angle $$\theta _{2}$$?

Answer 1

(a) A simple implication of Snell’s law is that $$\theta _{2}=\theta _{1}$$ when $$n_{1}=n_{2}$$. Since the angle of incidence is shown in figure (a) to be $$30^{0}$$, we look for a point in figure (b) where $$\theta _{2}=30^{0}$$. This seems to occur when $$n_{2}=1.7$$. By inference, then, $$n_{1}=1.7$$.
(b) From $$1.7\sin 60^{0}=2.4\sin \left ( \theta _{2} \right )$$ we get $$\theta _{2}=38^{0}$$.