Physics · Ch 10 — Wave Optics

Refraction of a Plane Wave

10.3.1

Refraction of a Plane Wave

Refraction of a Plane Wave Using Huygens Principle

Huygens principle states that every point on a wavefront acts as a source of secondary spherical wavelets. The new wavefront is the envelope of these wavelets. This principle is used here to derive the laws of refraction.

Step 1: Setting Up the Geometry

Consider a plane wavefront AB incident on the interface PP' separating medium 1 (speed $v_1$ ) and medium 2 (speed $v_2$ ).
The wavefront travels in the direction A'A, making an angle of incidence $i$ with the normal.
Let $t$ be the time taken for the wavefront to travel from B to C. In this time, the distance covered in medium 1 is:

BC = v_1 t

Step 2: Constructing the Refracted Wavefront

From point A (on the interface), draw a sphere of radius $v_2 t$ in medium 2. This sphere represents the secondary wavelet from A after time $t$ .
From point C (on the interface), draw a tangent plane to this sphere. The point of tangency is E.
The line CE is the refracted wavefront. The distance AE is:

AE = v_2 t

Step 3: Deriving Snell's Law

In triangle ABC (in medium 1), the angle at A is $i$ . Using the right triangle:

\sin i = \frac{BC}{AC} = \frac{v_1 t}{AC}

In triangle AEC (in medium 2), the angle at A is $r$ (angle of refraction). Using the right triangle:

\sin r = \frac{AE}{AC} = \frac{v_2 t}{AC}

Dividing the two equations gives:

\frac{\sin i}{\sin r} = \frac{v_1}{v_2}

This is the law of refraction in terms of wave speeds.

Step 4: Physical Interpretation

If $r < i$ (ray bends toward the normal), then $\sin i > \sin r$ , so $v_1 > v_2$ . This means the speed of light is less in the denser medium.
This prediction of wave theory is opposite to the corpuscular model and was confirmed by experiments.

Step 5: Introducing Refractive Indices

The refractive index of a medium is defined as:

n_1 = \frac{c}{v_1}, \quad n_2 = \frac{c}{v_2}

where $c$ is the speed of light in vacuum.

Substituting into the ratio:

\frac{v_1}{v_2} = \frac{n_2}{n_1}

Therefore, the law becomes Snell's law:

n_1 \sin i = n_2 \sin r

Step 6: Wavelength and Frequency

Let $\lambda_1$ and $\lambda_2$ be the wavelengths in medium 1 and medium 2 respectively. …

Figure 10.4A plane wave AB is incident at an angle i on the surface PP′ separating medium 1 and medium 2. The plane wave undergoes refraction and CE represents the refracted wavefront. The figure corresponds to v2 < v1 so that the refracted waves bends towards the normal.

$Fig. 10.4 — A plane wave AB is incident at an angle i on the surface PP′ separating medium 1 and medium 2. The plane wave undergoes refraction and CE represents the refracted wavefront. The figure corresponds to v2 < v1 so that the refracted waves bends towards the normal.$

Drawn by us to help you understand the concept clearly, and verified to make sure it's accurate. For exams, practice from your NCERT textbook's own diagram.

The figure shows a plane wavefront $AB$ approaching a slanted interface $PP'$ from medium 1 (above, with wave speed $v_1$ ) toward medium 2 (below, with wave speed $v_2$ ). The interface is drawn as a straight line, and a normal (dashed line) is drawn perpendicular to it at point $A$ . The incident wavefront $AB$ makes an angle $i$ with the interface — this is the angle of incidence.

Point $B$ is the last point of the incident wavefront to touch the interface, reaching point $C$ after time $\tau$ .
During this same time $\tau$ , a secondary wavelet (Huygens wavelet) spreads from point $A$ into medium 2 with speed $v_2$ , forming a sphere of radius $AE = v_2 \tau$ .
The refracted wavefront is the common tangent $CE$ drawn from $C$ to that sphere. It makes an angle $r$ with the interface — the angle of refraction.

Two right triangles are highlighted:

$\triangle ABC$ : right-angled at $B$ , with $BC = v_1 \tau$ and $AC$ as the hypotenuse.
$\triangle AEC$ : right-angled at $E$ , with $AE = v_2 \tau$ and $AC$ as the common hypotenuse.

Because $v_2 < v_1$ (medium 2 is denser), the refracted wavefront $CE$ is bent toward the normal, so $r < i$ .

Key formulas derived from this figure

From $\triangle ABC$ :

\sin i = \frac{BC}{AC} = \frac{v_1 \tau}{AC}

From $\triangle AEC$ :

\sin r = \frac{AE}{AC} = \frac{v_2 \tau}{AC}

Dividing these gives the refraction law:

$\frac{\sin i}{\sin r} = \frac{v_1}{v_2}$ …

Start your 14-day free trial to unlock the full solution →

← Previous · 10.3Refraction and Reflection of Plane Waves using Huygens Principle Next · 10.3.2 →Refraction at a Rarer Medium