In above figure (a), a beam of light in material 1 is incident on a boundary at an angle of $θ_{1}=30_{0}$. 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 $θ_{2}$ versus $n_{2}$ for a range of possible $n_{2}$ values. The vertical axis scale is set by $θ_{2a}=20_{0}$ and $θ_{2b}=40_{0}$. (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 $θ_{2}$?

Hard

Open in App

Solution

Verified by Toppr

(a) A simple implication of Snell’s law is that $θ_{2}=θ_{1}$ 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 $θ_{2}=30_{0}$. This seems to occur when $n_{2}=1.7$. By inference, then, $n_{1}=1.7$.

(b) From $1.7sin60_{0}=2.4sin(θ_{2})$ we get $θ_{2}=38_{0}$.