10.3 Refraction and Reflection of Plane Waves using Huygens Principle

10.3.1 Refraction of a plane wave

We will now use Huygens principle to derive the laws of refraction. Let PP ${ }^{\prime}$ represent the surface separating medium 1 and medium 2 , as shown in Fig. 10.4. Let $v_1$ and $v_2$ represent the speed of light in medium 1 and medium 2, respectively. We assume a plane wavefront $A B$ propagating in the direction A’A incident on the interface at an angle $i$ as shown in the figure. Let $\tau$ be the time taken by the wavefront to travel the distance BC.
Thus,$\mathrm{BC}=v_1 \tau$

In order to determine the shape of the refracted wavefront, we draw a sphere of radius $v_2 \tau$ from the point $A$ in the second medium (the speed of the wave in the second medium is $v_2$ ). Let CE represent a tangent plane drawn from the point C on to the sphere. Then, $\mathrm{AE}=v_2 \tau$ and $C E$ would represent the refracted wavefront. If we now consider the triangles ABC and AEC , we readily obtain
$\sin i=\frac{\mathrm{BC}}{\mathrm{AC}}=\frac{v_1 \tau}{\mathrm{AC}} \dots(10.1)$
$\begin{aligned} &\text { and }\\ &\sin r=\frac{\mathrm{AE}}{\mathrm{AC}}=\frac{v_2 \tau}{\mathrm{AC}} \dots(10.2) \end{aligned}$
where $i$ and $r$ are the angles of incidence and refraction, respectively. Thus we obtain
$\frac{\sin i}{\sin r}=\frac{v_1}{v_2} \dots(10.3)$
From the above equation, we get the important result that if $r<i$ (i.e., if the ray bends toward the normal), the speed of the light wave in the second medium $\left(v_2\right)$ will be less then the speed of the light wave in the first medium $\left(v_1\right)$. This prediction is opposite to the prediction from the corpuscular model of light and as later experiments showed, the prediction of the wave theory is correct. Now, if $c$ represents the speed of light in vacuum, then,
$n_1=\frac{c}{v_1} \dots(10.4)$
$\begin{aligned} &\text { and }\\ &n_2=\frac{c}{v_2} \dots(10.5) \end{aligned}$
are known as the refractive indices of medium 1 and medium 2, respectively. In terms of the refractive indices, Eq. (10.3) can be written as
$n_1 \sin i=n_2 \sin r \dots(10.6)$
This is the Snell’s law of refraction. Further, if $\lambda_1$ and $\lambda_2$ denote the wavelengths of light in medium 1 and medium 2, respectively and if the distance BC is equal to $\lambda_1$ then the distance AE will be equal to $\lambda_2$ (because if the crest from B has reached C in time $\tau$, then the crest from A should have also reached $E$ in time $\tau$ ); thus,
$\frac{\lambda_1}{\lambda_2}=\frac{\mathrm{BC}}{\mathrm{AE}}=\frac{v_1}{v_2} \dots(10.7)$
$\begin{aligned} &\text { or }\\ &\frac{v_1}{\lambda_1}=\frac{v_2}{\lambda_2} \dots(10.7) \end{aligned}$
The above equation implies that when a wave gets refracted into a denser medium ( $v_1>v_2$ ) the wavelength and the speed of propagation decrease but the frequency $v(=v / \lambda)$ remains the same.

10.3.2 Refraction at a rarer medium

We now consider refraction of a plane wave at a rarer medium, i.e., $v_2>v_1$. Proceeding in an exactly similar manner we can construct a refracted wavefront as shown in Fig. 10.5. The angle of refraction will now be greater than angle of incidence; however, we will still have $n_1 \sin i=n_2 \sin r$. We define an angle $i_c$ by the following equation
$\sin i_c=\frac{n_2}{n_1} \dots(10.8)$
Thus, if $i=i_c$ then $\sin r=1$ and $r=90^{\circ}$. Obviously, for $i>i_c$, there can not be any refracted wave. The angle $i_c$ is known as the critical angle and for all angles of incidence greater than the critical angle, we will not have any refracted wave and the wave will undergo what is known as total internal reflection. The phenomenon of total internal reflection and its applications was discussed in Section 9.4.

10.3.3 Reflection of a plane wave by a plane surface

We next consider a plane wave AB incident at an angle $i$ on a reflecting surface MN. If $v$ represents the speed of the wave in the medium and if $\tau$ represents the time taken by the wavefront to advance from the point $B$ to $C$ then the distance
$\mathrm{BC}=v \tau$
In order to construct the reflected wavefront we draw a sphere of radius $v \tau$ from the point A as shown in Fig. 10.6. Let CE represent the tangent plane drawn from the point C to this sphere. Obviously
$\mathrm{AE}=\mathrm{BC}=v \tau$

If we now consider the triangles EAC and BAC we will find that they are congruent and therefore, the angles $i$ and $r$ (as shown in Fig. 10.6) would be equal. This is the law of reflection.

Once we have the laws of reflection and refraction, the behaviour of prisms, lenses, and mirrors can be understood. These phenomena were discussed in detail in Chapter 9 on the basis of rectilinear propagation of light. Here we just describe the behaviour of the wavefronts as they undergo reflection or refraction. In Fig. 10.7(a) we consider a plane wave passing through a thin prism. Clearly, since the speed of light waves is less in glass, the lower portion of the incoming wavefront (which travels through the greatest thickness of glass) will get delayed resulting in a tilt in the emerging wavefront as shown in the figure. In Fig. 10.7(b) we consider a plane wave incident on a thin convex lens; the central part of the incident plane wave traverses the thickest portion of the lens and is delayed the most. The emerging wavefront has a depression at the centre and therefore the wavefront becomes spherical and converges to the point F which is known as the focus. In Fig. 10.7(c) a plane wave is incident on a concave mirror and on reflection we have a spherical wave converging to the focal point F. In a similar manner, we can understand refraction and reflection by concave lenses and convex mirrors.

From the above discussion it follows that the total time taken from a point on the object to the corresponding point on the image is the same measured along any ray. For example, when a convex lens focusses light to form a real image, although the ray going through the centre traverses a shorter path, but because of the slower speed in glass, the time taken is the same as for rays travelling near the edge of the lens.

Example 10.1: (a) When monochromatic light is incident on a surface separating two media, the reflected and refracted light both have the same frequency as the incident frequency. Explain why?
(b) When light travels from a rarer to a denser medium, the speed decreases. Does the reduction in speed imply a reduction in the energy carried by the light wave?
(c) In the wave picture of light, intensity of light is determined by the square of the amplitude of the wave. What determines the intensity of light in the photon picture of light.

Solution: (a) Reflection and refraction arise through interaction of incident light with the atomic constituents of matter. Atoms may be viewed as oscillators, which take up the frequency of the external agency (light) causing forced oscillations. The frequency of light emitted by a charged oscillator equals its frequency of oscillation. Thus, the frequency of scattered light equals the frequency of incident light.
(b) No. Energy carried by a wave depends on the amplitude of the wave, not on the speed of wave propagation.
(c) For a given frequency, intensity of light in the photon picture is determined by the number of photons crossing an unit area per unit time.