Huygens’ Principle, Wavefront, Coherent Sources and Interference of Light

1. Definition of Wavefront

A wavefront is the locus of all points of a medium which are vibrating in the same phase at a given instant.

In simple words, all points on a wavefront have the same phase of vibration.

Examples:

• For a point source, the wavefront is spherical.

• For a very distant source, the wavefront is plane.

• For a line source, the wavefront is cylindrical.

2. Huygens’ Principle

Huygens’ principle states that every point on a given wavefront acts as a source of secondary wavelets. These secondary wavelets spread in the forward direction with the speed of light. The new wavefront at any later time is the common tangent or envelope of these secondary wavelets.

This principle helps us understand reflection, refraction, diffraction and interference of light.

3. Huygens’ Explanation of Diffraction

Diffraction is the bending or spreading of light waves around the edges of an obstacle or aperture.

According to Huygens’ principle, every point of the slit acts as a source of secondary wavelets. These wavelets spread in different directions. When the size of the slit is comparable to the wavelength of light, the spreading becomes clearly visible. This spreading of light is called diffraction.

Diffraction is more pronounced when:

Size of aperture ≈ wavelength of light

That is why sound diffraction is easily observed, but light diffraction is not easily observed in daily life because the wavelength of light is very small.

4. Definition of Coherent Sources

Two sources are called coherent sources if they emit light waves of the same frequency or same wavelength and maintain a constant phase difference with time.

For sustained interference, coherent sources are necessary.

Conditions for coherent sources:

• Same frequency

• Same wavelength

• Constant phase difference

• Same state of polarization

• Nearly equal intensities for clear fringes

5. Interference of Light

Interference of light is the phenomenon in which two coherent light waves superpose and redistribute energy in space, producing alternate bright and dark fringes.

When two waves meet at a point, their displacements add according to the principle of superposition.

If the waves meet in the same phase, brightness increases.
If the waves meet in opposite phase, darkness is produced.

6. Constructive Interference

Constructive interference occurs when two coherent waves meet in the same phase.

Condition for constructive interference:

Path difference = n lambda

where n = 0, 1, 2, 3, …

At these points, bright fringes are obtained.

So,

Bright fringe condition:

Delta x = n lambda

7. Destructive Interference

Destructive interference occurs when two coherent waves meet in opposite phase.

Condition for destructive interference:

Path difference = (2n – 1) lambda / 2

where n = 1, 2, 3, …

At these points, dark fringes are obtained.

So,

Dark fringe condition:

Delta x = (2n – 1) lambda / 2

8. How Bright and Dark Fringes are Formed

In Young’s double slit experiment, two narrow slits act as coherent sources. Light waves from these two slits overlap on the screen.

At some points on the screen, path difference is an integral multiple of wavelength. These points become bright.

At some other points, path difference is an odd multiple of half wavelength. These points become dark.

For bright fringe:

Path difference = n lambda

For dark fringe:

Path difference = (2n – 1) lambda / 2

This is how alternate bright and dark fringes are formed on the screen.

9. Meaning of Missing Wavelength

Missing wavelength means that a particular wavelength is absent at a particular point because destructive interference occurs there.

In simple words:

Missing wavelength means dark fringe for that wavelength.

If a wavelength is missing at a point, it means the intensity of that wavelength at that point is minimum or zero.

So, for missing wavelength, we always apply the dark fringe condition:

Path difference = (2n – 1) lambda / 2

10. Applications of Interference in Daily Life

Interference is used in many practical situations:

1. Anti-reflection coating on lenses

2. Thin film colours in soap bubbles

3. Colours in oil films on water

4. Testing flatness of glass plates

5. Michelson interferometer

6. Measuring wavelength of light

7. Measuring very small thickness

8. Optical instruments

9. Holography

10. Interference filters

11. Important Question: Missing Wavelength in Young’s Double Slit Experiment

Question

When light is used to illuminate the two slits of Young’s experiment, the separation between the slits is b, and the screen is at a distance D, where D is much greater than b. At a point directly in front of one of the slits, certain wavelengths are missing. Find the missing wavelengths.

Diagram Explanation

Let the two slits be S1 and S2.

The separation between the slits is:

S1S2 = b

The screen is at distance:

D

Point P is directly in front of one slit, say S1.

So, the distance of point P from the central line is:

x = b / 2

Because the central line lies midway between the two slits.

Simple diagram:

S1 •
        |
        |——– P
        |
S2 •

Here,

S1S2 = b

Point P is directly in front of S1.

Therefore,

x = b / 2

12. Formula Used

For Young’s double slit experiment, path difference is:

Delta x = x b / D

For dark fringe:

Delta x = (2n – 1) lambda / 2

Now put:

x = b / 2

So,

Delta x = (b / 2) × b / D

Delta x = b² / 2D

Now apply dark fringe condition:

b² / 2D = (2n – 1) lambda / 2

Multiply both sides by 2:

b² / D = (2n – 1) lambda

Therefore,

lambda = b² / [(2n – 1)D]

Final Answer

Missing wavelengths are:

lambda = b² / [(2n – 1)D]

where,

n = 1, 2, 3, 4, …

So the missing wavelengths are:

For n = 1:

lambda = b² / D

For n = 2:

lambda = b² / 3D

For n = 3:

lambda = b² / 5D

For n = 4:

lambda = b² / 7D

Therefore, the missing wavelengths are:

b²/D, b²/3D, b²/5D, b²/7D, …

Kumar Sir Style Final Concept

Always remember: whenever the question says “wavelength is missing”, it means that wavelength is producing a dark fringe at that point. So we must apply the destructive interference condition.

At the point directly in front of one slit, the distance from central maximum is b/2. Put x = b/2 in the path difference formula and then apply the dark fringe condition. This gives the missing wavelengths:

lambda = b² / [(2n – 1)D]