Name: Primary and Secondary Wavefront
Uploaded: 2019-07-19
Duration: 2 min 30 s
Description: The creation of wavefront causes a “chain reaction” of creating more wavefronts that pave the way for the propagation of waves. Let's dive deep into this phe...

Question

Using Huygens&rsquo;s wave theory, derive Snell&rsquo;s law of refraction.

Accepted Answer

Consider a plane wave front AB incident on the surface PQ separating those two media named 1 and 2.

Let the medium 1 is a rarer medium of refractive index $n_{1}$ in which light travels with a velocity $c_{1}$ . The medium 2 is denser medium of refractive index $n_{2}$ in which light travels with a velocity $c_{2}$ .

The angle between the incident ray $F A$ and the normal $N A$ at the point of incidence A is the angle of incidence.

The angle between the incident plane is also equal to the angle between wave front $B A$ and the surface of separation PQ.

So $∠ B A D$ is the angle of incidence of the incident plane wave front $A B$ .

Like this, the angle between the wave front which is refracted and the surface of separation $P Q$ is equal to the angle of refraction.

That is, $∠ A D C = r$ .

Consider the triangles $B A D$ and $A C D$ figure above.

The sine of the angle $B A D$ can be found by taking the ratio between the opposite sides to the adjacent side.

That is,
$\sin i = \sin ∠ B A D = \frac{B D}{A D} = \frac{c_{1} t}{A D}$ ---------- (1)
Similarly,
$\sin r = \sin ∠ A D C = \frac{A C}{A D} = \frac{c_{2} t}{A D}$ ----------- (2)

Dividing equation (2) by equation (1),
$\frac{\sin i}{\sin r} = \frac{c_{1}}{c_{2}}$
This is equal to ratio of refractive index of the second medium (2) to the first medium (1).
$\frac{c_{1}}{c_{2}} = \frac{n_{2}}{n_{1}} \frac{\sin i}{\sin r} = \frac{n_{2}}{n_{1}}$
This equation proves Snell's law

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