Huygens principle

Huygens’ Principle is a key idea in physics explaining how waves travel along a material. It was derived in the 17th century by a Dutch scientist called Christiaan Huygens who simply supposed that every point on an advancing wavefront acts as a source of secondary waves which move out in all directions. The totality of these wavelets gives rise to the new position of the wavefront at a later instant according to Huygens’s principle. Through this principle, optics, acoustics as well as other fields needed to interpret phenomena like interference or diffraction can be better understood.

In this article, we will discuss the concept of Huygens Principle. All points on a wavefront can be considered as secondary spherical wavelets sources influencing the waveform shape later. Over the last ten years of the JEE Main exam (from 2013 to 2023), a total of two questions have been asked on this concept.

Huygens Principle

According to the Huygens principle , Every point on the given wavefront acts as a source of a new disturbance called secondary wavelets. And a common tangent to these secondary wavelets in the forward direction at any instant gives the new wavefront at that instant as shown in the below figure. This is called secondary wavefront.

The Lightwave will follow the Laws of Reflection. Let's understand this with the help of the Huygens principle.

Consider a plane wavefronts travels towards a plane reflecting surface as shown in the figure.

Let $A B$ and $C D$ as the incident and reflected wavefronts respectively.
Let at $\mathrm{t}=0$ wave is at A and at $t=\tau$ wave is at C .
if $\mathrm{v}_1$ is the velocity of the wave $B C=A D=v_1 \tau$
And as $\triangle A B C \cong \triangle A D C$ So we get $i=r$
This verifies the first law of reflection which states that the angle of incidence i and angle of reflection $r$ are always equal.
Similarly from the figure, we can say that the incident wavefront, the reflected wavefront and normal lie in the same plane.
This again verifies the second law of reflection.
Therefore, the two laws of Reflection are verified using Huygens's Principle.
The Lightwave will follow the Laws of Refraction. Let's understand this with the help of the Huygens principle.

Consider a plane wavefronts travels towards a plane $A C$ as shown in the above figure.
Let $A B$ and $C D$ as the incident and refracted wavefronts respectively.
Let at $\mathrm{t}=0$ wave is at A and at $t=\tau$ wave is at C .
if $\mathrm{v}_1$ is the velocity of the wave in medium 1 then $B C=v_1 \tau$
similarly $\mathrm{v}_2$ is the velocity of the wave in the medium 2 then $A D=v_2 \tau$
For $\triangle A B C \rightarrow \sin (i)=\frac{B C}{A C}=\frac{v_1 \tau}{A C}$
similarly
For $\triangle A C D \rightarrow \sin (r)=\frac{A D}{A C}=\frac{v_2 \tau}{A C}$
So we get $\frac{\sin (i)}{\sin (r)}=\frac{v_1 \tau}{v_2 \tau}=\frac{v_1}{v_2}$
And we know $\mu \quad \alpha \quad v$
So we get

$
\frac{\sin (i)}{\sin (r)}=\frac{v_1}{v_2}=\frac{\mu_2}{\mu_1}=\mu_{21}=\text { constant }
$
This verifies the first law of refraction.

Similarly from the figure, we can say that the incident wavefront, the refracted wavefront and normal lie in the same plane.

This again verifies the second law of Refraction.

Therefore, the two laws of Refraction are verified using Huygens's Principle.

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Solved Examples Based on Huygens’ Principle

Example 1: A plane surface separating medium 1 and medium 2 has refractive indexes as $\mu_1$ and $\mu_2$ respectively. (where $\mu_1<\mu_2$ ). Let the with the plane surface. So the relation between angle $i$ and $r$ for the plane wavefront approaching a plane surface in medium 1 is given as :
1) $\frac{\sin (i)}{\sin (r)}=\frac{\mu_2}{\mu_1}$
2) $\frac{\sin (i)}{\sin (r)}=\mu_2-\mu_1$
3) $\frac{\sin (i)}{\sin (r)}=\mu_2$

4) $\frac{\sin (i)}{\sin (r)}=\mu_1$

Solution:
The Lightwave follows the Laws of Refraction. Using Snell's Law :

$
\frac{\sin (i)}{\sin (r)}=\frac{\mu_2}{\mu_1}
$
Hence, the answer is the option (1).

Example 2: On a hot summer night, the refractive index of air is smallest near the ground and increases with height from the ground. When a light beam is directed horizontally, the Huygens’ principle leads us to conclude that as it travels, the light beam :

1) becomes narrower

2) goes horizontally without any deflection

3) bends downwards

4) bends upwards

Solution:

Refraction of plane wave using Huygens principle -

If v₁ is the velocity of the wave in medium 1

similarly, $\mathrm{v}_2$ is the velocity of the wave in the medium 2 then

$
\frac{\sin (i)}{\sin (r)}=\frac{v_1}{v_2}=\frac{\mu_2}{\mu_1}=\mu_{21}=\text { constant }
$
Consider a plane wavefront travelling horizontally. As the refractive index of air increases with height. So the speed of the wavefront decreases with height.

So the value of r is less than i.

That refracted ray would bend towards the normal

Hence the light beam bends upwards.

Hence, the answer is the option 4.

Example 3: If the wave gets refracted into a denser medium, then which of the following is true?

1) Wavelength, speed and frequency decreases.

2) Wavelength increases, speed decreases and frequency remains constant..

3) Wavelength and speed decreases but frequency remains constant.

4) Wavelength,speed and frequency increases.

Solution:

$\begin{aligned} & \mathrm{V}=\lambda \mathrm{f} \\ & \mathrm{V}=\frac{\mathrm{C}}{\mu}\end{aligned}$
When the wave goes from rarer to denser medium speed decreases, the frequency remains the same, and therefore wavelength decreases.

The number of waves per unit from one medium to another medium remains constant to maintain continuity.
Hence, the answer is the option (3).

Example 4: A light wave travels from medium 1 with a velocity of $2 \times 10^8 \mathrm{~m} / \mathrm{s}$ to medium 2 with a velocity of $1.5 \times 10^8 \mathrm{~m} / \mathrm{s}$. If the angle of incidence is 30 degrees, what is the angle of refraction?
1) 5 degrees
2) 10 degrees
3) 22 degrees
4) 60 degrees

Solution:
Using the formula

$
\frac{\sin (i)}{\sin (r)}=\frac{v_1}{v_2}
$

we can find

$
\sin (\mathrm{r})=\frac{v_2}{v_1} \quad \times \sin (\mathrm{i})
$

Substituting values, $\sin (r)=\left(\left(1.5 \times 10^8 \mathrm{~m} / \mathrm{s}\right) /\left(2 \times 10^8 \mathrm{~m} / \mathrm{s}\right)\right) \times \sin (30)=0.75 \times 0.5=0.375$
Taking the inverse sine

we get the angle of refraction as approximately 22 degrees.

Hence, the answer is the option (3).

Example 5: A light wave is incident from air to a medium with a refractive index of 1.5. If the angle of incidence is 45 degrees, what is the angle of refraction?

1) 30 degrees

2) 45 degrees

3) 60 degrees

4) 75 degrees

Solution:

We know that

$
\frac{\sin (i)}{\sin (r)}=\frac{n_2}{n_1}
$

where $n_1$ is the refractive index of air (approximately 1 ) and $\mathrm{n}_2$ is the refractive index of the medium ( 1.5 in this case).

Substituting values,

$
\sin (r)=\frac{1}{1.5} \times \sin (45)=\sin (r)=0.47
$
Taking the inverse sine, we get the angle of refraction as 30 degrees.

Hence, the answer is the option (1).

Summary

According to Huygens’ Principle, every point on a wavefront is treated as a secondary source of wavelets that spread in all directions. Hence, combing up to form the new wavefront some time later thereafter they had been emitted! For instance let me introduce us to some of these basic facts about how do we study light propagation using this principle; it’s one way among many others which help us comprehend why we see wavelengths emerging from different colors but they still appear white while others are black (or invisible). The phenomenon of reflection – when a wave bounces off something hard like glass or metal – can also be explained using the same principle. Additionally, this happens for electric forces between charges which then creates an effect known as electrical waves plus magnetic ones.