# Refraction of a normal incident ray

From the view of law of refraction $$\sin\theta_1/\sin\theta_2=v_1/v_2$$

how can one justify the case that the incident ray enter vertical (normal to the boundary, i.e. $$\theta_1=0$$)?

If the law is reformulated into $$v_2 \sin \theta_1 = v_1 \sin \theta_2$$ the division by $$0$$ is avoided and you get $$\theta_2 = 0$$ for all possible $$v_2$$.

• But is this true according to physics experiments?The light does not refract in this normal case? Dec 21, 2020 at 23:04
• @AliTaghavi Yes, if it propagates along the surface normal on one side it will continue to do so on the other side. Think about the rotational symmetry around the propagation axis. It would be strange if the symmetry would be broken by the refraction. Dec 21, 2020 at 23:10

This is how you should prove a statement. One can prove it from Fermat's principle from the figure above, it states that light travels between two points along the path that requires the least time, as compared to other nearby paths.

$$t=\frac{\sqrt{a^2+x^2}}{v}+\frac{\sqrt{b^2+(d-x)^2}}{v'}$$

above statement tells us

$$\frac{dt}{dx}=\frac{x}{v\sqrt{a^2+x^2}}-\frac{d-x}{v'\sqrt{b^2+(d-x)^2}}=0$$

this is equivalent with

$$\frac{\sin(\theta_{1})}{v}=\frac{\sin(\theta_{2})}{v'}$$

If your incident ray enters normal to the surface $$x=0$$, again write the same principle (keep it general)

$$t=\frac{\sqrt{a^2+x^2}}{v}+\frac{\sqrt{b^2+(d-x)^2}}{v'}$$

thus $$x=0$$

$$\frac{dt}{dx}=0-\frac{d}{v'\sqrt{b^2+d^2}}=0$$

and this is possible if $$d=0 \rightarrow d-x=0$$.