Question

Using Huygen's concept of wave front, derive Snell's law of refraction.

Answer 1

Here we must take into account a different speed of light in the upper and lower media. If the speed of light in vacuum is $c$, we express the speed in the upper medium by the ratio $\frac{c}{n_i}$, where $n_i$ is the refractive index. Similarly, the speed of light in the lower medium is $\frac{c}{n_t}$. The points $D$, $E$ and $F$ on the incident wavefront arrive at points $D$, $J$ and $I$ of the plane interface $XY$ at different times. In the absence of the refracting surface, the wavefront $GI$ is formed at the instant ray $DF$ reaches $I$. During the progress of ray $CF$ from $F$ to $I$ in time $t$, however, the ray $AD$ has entered the lower medium, where the speed is different. Thus if the distance $DG$ is $v_{i}t$, a wavelet of radius $v_{t}t$ is constructed with center at $D$. The radius $DM$ can also be expressed as

$DM=v_{t}t=v_t\left(\frac{DG}{v_i}\right)=\left(\frac{n_i}{n_t}\right)\times DG$

Similarly, a wavelet of radius $\frac{n_i}{n_t}JH$ is drawn centered at $J$. The new wavefront $KI$ includes point $I$ on the interface and is tangent to the two wavelets at points $M$ and $N$. The geometric relationship between the angles ${\theta }_i$ and ${\theta }_t$, formed by the representative incident ray $AD$ and refracted ray $DL$, is Snell's law, which may be expressed as

$n_{i}sin{\theta }_i=n_{t}sin{\theta }_t$