Introduction

Q: How does Huygens' principle prove Snell's law?

While one end of an incident wavefront travels BC = v1 t in the first medium, the wavelet from the other end grows to AE = v2 t in the second. Since BC = AC sin i and AE = AC sin r, equating the times gives sin i / sin r = v1 / v2 = n21, which is Snell's law.

Q: What is the ratio of maximum to minimum intensity in interference?

For waves of intensities I1 and I2, Imax/Imin = (root I1 + root I2) squared divided by (root I1 minus root I2) squared, which equals ((a1 + a2)/(a1 - a2)) squared in terms of amplitudes. For intensities in the ratio 4:1 the amplitudes are 2:1 and the ratio is 9:1.

PhysicsWave OpticsFor JEE aspirants

Wave optics studies light as a wave, which explains effects that ray optics cannot: interference, diffraction and polarisation. Its two starting tools are the wavefront (a surface of equal phase) and Huygens' principle (every point on a wavefront is a source of secondary wavelets). From them come the laws of reflection and refraction and the interference formula $I = I_{1} + I_{2} + 2 I_{1} I_{2} cos Δ ϕ$ . Wave optics carries steady marks in JEE Main and NEET, and this introduction is the base for YDSE, thin films and diffraction.

On this page1Light as an EM wave2Wavefronts3Huygens' principle4Reflection and refraction5Coherent sources6Adding two waves7Interference and intensity

Key Formulas - Quick Reference

Ray $⊥$ wavefront. Intensity: spherical $I \propto \frac{1}{r ^{2}}$ , cylindrical $I \propto \frac{1}{r}$ , plane $I$ constant (amplitude $A \propto I$ ).
Snell's law from Huygens: $\frac{sin i}{sin r} = \frac{v _{1}}{v _{2}} = \frac{n _{2}}{n _{1}} = n_{21}$
In a medium of index $n$ : $v = \frac{c}{n}$ , $λ_{n} = \frac{λ}{n}$ , frequency unchanged.
Phase and path difference: $Δ ϕ = \frac{2 π}{λ} Δ x$
Resultant amplitude: $A^{2} = a_{1}^{2} + a_{2}^{2} + 2 a_{1} a_{2} cos ϕ$ , $tan ϕ_{0} = \frac{a _{2} sin ϕ}{a _{1} + a _{2} cos ϕ}$
Resultant intensity: $I = I_{1} + I_{2} + 2 I_{1} I_{2} cos Δ ϕ$ ; for $I_{1} = I_{2} = I_{0}$ : $I = 4 I_{0} cos^{2} \frac{Δ ϕ}{2}$
Constructive: $Δ x = nλ$ , $Δ ϕ = 2 nπ$ , $I_{ma x} = (I_{1} + I_{2})^{2}$
Destructive: $Δ x = (2 n + 1) \frac{λ}{2}$ , $Δ ϕ = (2 n + 1) π$ , $I_{min} = (I_{1} - I_{2})^{2}$
$\frac{I _{ma x}}{I _{min}} = (\frac{a _{1} + a _{2}}{a _{1} - a _{2}})^{2}$ with $\frac{a _{1}}{a _{2}} = \frac{I _{1}}{I _{2}}$
Incoherent sources: $I = I_{1} + I_{2}$ ( $n$ sources: $n I_{0}$ ); coherent, in phase: $n^{2} I_{0}$

1. Light as an Electromagnetic Wave

Light is a transverse electromagnetic (EM) wave: an oscillating electric field $E$ and magnetic field $B$ , perpendicular to each other and to the direction of travel. Visible light is the small part of the EM spectrum that our eyes detect, from about $400 nm$ (violet) to $700 nm$ (red).

Figure 1: The electromagnetic spectrum. Visible light is a narrow band (

400

700 nm

) whose wavelength is far smaller than everyday objects, which is why ray optics usually works.

Huygens proposed the wave theory in 1678. It was accepted after Young's double-slit experiment (1801) and Foucault's measurement (1850) that light is slower in water, as the wave theory predicts and the corpuscular theory does not. Maxwell later showed that light is an EM wave.

Figure 2: Light is a transverse EM wave.

E

(along

y

) and

B

(along

z

) oscillate in phase, perpendicular to each other and to the direction of travel; their amplitudes obey

E_{0} = c B_{0}

and the wave moves at

c = 3 \times 1 0^{8} m s^{- 1}

in vacuum.

The two fields rise and fall together (in phase), and their amplitudes are linked by $E_{0} = c B_{0}$ . Because the vibration is perpendicular to the direction of travel, light is a transverse wave, which is why it can be polarised (see the Diffraction page).

1.1 Ray optics or wave optics?

Which model we use depends on the size of the obstacle or opening compared with the wavelength $λ$ .

Feature	Geometrical (ray) optics	Wave optics
When valid	Object or aperture size $≫ λ$	Object or aperture size comparable to $λ$
Light is treated as	Rays travelling in straight lines	Waves with amplitude and phase
Explains	Reflection, refraction, images by mirrors and lenses	Interference, diffraction, polarisation (and reflection, refraction)
Key idea	Laws of reflection and Snell's law	Wavefront, Huygens' principle, superposition

Rule of thumb: ray optics is the limit $λ \to 0$ of wave optics. If a question gives a slit or obstacle a few micrometres wide with visible light, expect wave effects; if it gives a lens of a few centimetres, ray optics is enough.

2. Wavefront

★ Must learnWavefront: the locus of all points that vibrate in the same phase at a given instant.

The direction of propagation (the ray) is always perpendicular to the wavefront.
The distance between two successive wavefronts that differ in phase by $2 π$ is one wavelength $λ$ .
Every point of a wavefront acts as a new source of secondary wavelets (Huygens).
Points on one wavefront have zero phase difference, so any two of them act as coherent sources.

2.1 Shapes of wavefronts

The shape depends on the source. Energy from a point source spreads over a growing sphere, from a line source over a growing cylinder, and a very distant source gives flat (plane) wavefronts.

Figure 3: Wavefront shape depends on the source. Rays (arrows) are always perpendicular to the wavefront; energy spreads over

4 π r^{2}

(sphere),

2 π r l

(cylinder) or not at all (plane).

Source	Wavefront	Intensity $I$	Amplitude $A$
Point source (small bulb)	Spherical	$\propto \frac{1}{r ^{2}}$	$\propto \frac{1}{r}$
Line source (slit, tube light)	Cylindrical	$\propto \frac{1}{r}$	$\propto \frac{1}{r}$
Very distant source (Sun, star), or a point source at the focus of a convex lens	Plane	constant	constant

Why these laws? Power $P$ is shared over the wavefront area: $I = \frac{P}{4 π r ^{2}}$ for a sphere and $I = \frac{P}{2 π r l}$ for a cylinder of length $l$ . Because $I \propto A^{2}$ , the amplitude falls as $I$ .

Key idea

Rays are always perpendicular to wavefronts, and the wavefront's shape tells you the source: a sphere for a point, a cylinder for a line, a plane for a very distant source.

3. Huygens' Principle

Huygens' principle is a geometrical method to find the position of a wavefront at a later time from its position now.

Every point on a given wavefront (the primary wavefront) acts as a fresh source of secondary wavelets.
The secondary wavelets spread in all directions with the speed of light in that medium, $v$ .
After a time $t$ each wavelet is a sphere of radius $v t$ .
The forward envelope (common tangent) of these wavelets is the new wavefront at time $t$ .
There is no backward wavefront: the backward envelope is ignored (Kirchhoff later justified this with a direction factor that is zero backwards).

Figure 4: Huygens' construction. Each point on the old wavefront (dots) sends out a wavelet of radius

v t

in time

t

, and the forward envelope is the new wavefront. The backward envelope (grey dashes) does not exist.

Two useful consequences follow. Every ray takes the same time to go from one wavefront to the next, and a plane wavefront in a uniform medium stays plane while a spherical one stays spherical with a larger radius.

4. Refraction and Reflection by Huygens' Principle

4.1 Refraction of a plane wave (Snell's law)

A plane wavefront AB meets the boundary between medium 1 (speed $v_{1}$ ) and medium 2 (speed $v_{2}$ ) at angle $i$ . The angle between the wavefront and the surface equals the angle between the ray and the normal.

Figure 5: Refraction into a slower medium, drawn exactly for

i = 5 0^{\circ}

and

n_{21} = 1.5

(so

r = 30. 7^{\circ}

). While B travels

B C = v_{1} t

, the wavelet from A grows to

A E = v_{2} t

; the wavefront bends towards the normal.

End A touches the surface first. End B still has to travel $B C$ in medium 1, taking time $t = \frac{B C}{v _{1}}$ .
In right triangle $A B C$ : $B C = A C sin i$ , so $t = \frac{A C sin i}{v _{1}}$ .
In the same time the wavelet from A grows inside medium 2 to radius $A E = v_{2} t$ .
The tangent CE from C to this wavelet is the refracted wavefront (the wavelet from any point D lying between A and C also touches CE). In right triangle $A E C$ : $A E = A C sin r$ , so $t = \frac{A C sin r}{v _{2}}$ .
Equate the two times:
$\frac{A C sin i}{v _{1}} = \frac{A C sin r}{v _{2}} \Rightarrow \frac{sin i}{sin r} = \frac{v _{1}}{v _{2}}$

★ Must learnSnell's law (wave form):

\frac{sin i}{sin r} = \frac{v _{1}}{v _{2}} = \frac{n _{2}}{n _{1}} = n_{21}

, where

n_{1} = \frac{c}{v _{1}}

and

n_{2} = \frac{c}{v _{2}}

are absolute refractive indices.

If $v_{2} < v_{1}$ (light enters a denser medium), then $r < i$ and the ray bends towards the normal. For water $n = 1.33$ , so $v = \frac{c}{1.33} \approx 0.75 c$ . Foucault's measurement of this lower speed was a decisive test of the wave theory.

4.2 What changes and what does not

The wavefronts on both sides of the boundary must match along the surface, so the number of waves arriving per second equals the number leaving: the frequency does not change. Speed and wavelength change together:

ν = \frac{v _{1}}{λ _{1}} = \frac{v _{2}}{λ _{2}} \Rightarrow \frac{λ _{2}}{λ _{1}} = \frac{v _{2}}{v _{1}} = \frac{n _{1}}{n _{2}}, λ_{m e d i u m} = \frac{λ _{v a c uu m}}{n}

Figure 6: Crossing into glass (

n = 1.5

), each crest must stay matched at the boundary, so the frequency stays

ν = 5 \times 1 0^{14} Hz

while

v

and

λ

both fall by

n

λ = 600 \to 400 nm

(Solved Example 6). The smaller amplitude in glass is the transmitted fraction

\frac{2}{1 + n} = 0.8

Exam Trick

Frequency is the fingerprint. Colour is decided by frequency, so light keeps its colour in water. Only $v$ and $λ$ shrink by the factor $n$ . Energy of a photon $h ν$ is also unchanged.

4.3 Refraction into a rarer medium and total internal reflection

If $v_{2} > v_{1}$ the wavelet from A grows faster than B moves, so $r > i$ and the ray bends away from the normal. When $sin r = 1$ the refracted wave grazes the surface; this angle of incidence is the critical angle, $sin i_{c} = \frac{v _{1}}{v _{2}} = \frac{n _{2}}{n _{1}}$ . For larger $i$ no refracted wavefront can be drawn and total internal reflection occurs.

4.4 Reflection of a plane wave

Figure 7: Reflection by Huygens' principle (

i = 3 5^{\circ}

). The speed is the same before and after, so

A E = B C

; triangles

A B C

and

C E A

are congruent and

i = r

Now the wavelet from A stays in medium 1, so $A E = v t = B C$ . Right triangles $A B C$ and $C E A$ share the hypotenuse $A C$ and have $A E = B C$ , so they are congruent and $∠ i = ∠ r$ : the law of reflection.

4.5 Wavefronts through a prism, a lens and a mirror

Figure 8: Plane wavefront in, new shape out. The part of the wavefront that spends longer in glass (or travels further to the mirror) lags behind, so the prism tilts it and the lens and mirror make it converge to F.

Thin prism: the lower part of the wavefront crosses more glass (the base is thicker), so it is delayed more; the emerging wavefront tilts and the ray bends towards the base.
Convex lens: the centre of the wavefront crosses the thickest glass and lags most; a plane wavefront becomes a spherical wavefront converging to the focus $F$ .
Concave mirror: the centre of the wavefront travels further before and after reflection, so again a converging spherical wavefront forms.
Concave lenses and convex mirrors turn a plane wavefront into a diverging spherical one by the same time-delay argument.

JEE Advanced

Equal optical time. Between an object point and its image, every ray takes the same time (equal optical path $\sum n l$ ). A ray through the thick centre of a convex lens is shorter in air but longer in glass; the two effects balance exactly. This is why a lens forms a sharp image, and it is the same idea as Fermat's principle.

Key idea

On refraction the frequency stays fixed; speed and wavelength both shrink by the same factor

n

Quick Recall: tap to check

Which wavefront does a long tube light produce close by?

Cylindrical:

I \propto \frac{1}{r}

and

A \propto \frac{1}{r}

Light of

600 nm

enters glass (

n = 1.5

). What are its wavelength and frequency there?

λ = 400 nm

; the frequency is unchanged (

5 \times 1 0^{14} Hz

Why is the backward envelope of Huygens' wavelets ignored?

No backward wave is observed; Kirchhoff's direction factor is zero straight backwards.

In the Huygens proof of reflection, why are triangles ABC and CEA congruent?

Both are right-angled, share the hypotenuse

A C

, and

A E = B C = v t

5. Superposition and Coherent Sources

★ Must learnPrinciple of superposition: when two or more waves pass through a point at the same time, the net disturbance is the sum of the disturbances each wave would produce alone:

y = y_{1} + y_{2} + \dots

The "disturbance" is displacement for a wave on a string, pressure change for sound and the electric field $E$ for light. When two light waves travelling in almost the same direction superpose, the intensity is redistributed in space: bright and dark regions appear. This redistribution is interference.

5.1 Coherent and incoherent sources

Coherent sources keep a constant phase difference with time (and have the same frequency). Only coherent sources give a steady (sustained) interference pattern.

	Coherent sources	Incoherent sources
Phase difference	Constant in time	Changes randomly (about every $1 0^{- 10} s$ for ordinary sources)
Intensities add as	$I = I_{1} + I_{2} + 2 I_{1} I_{2} cos Δ ϕ$	$I = I_{1} + I_{2}$ (because $⟨ cos Δ ϕ ⟩ = 0$ )
Pattern	Steady bright and dark fringes	Uniform illumination, no fringes
Examples	Two slits lit by one source, source and its mirror image, laser beams	Two separate bulbs, two halves of a sodium lamp

Figure 9: Two equal sources,

I_{1} = I_{2} = I_{0}

. A fixed

Δ ϕ

(

\frac{π}{3}

here) gives a steady

I = 2 I_{0} (1 + cos Δ ϕ) = 3 I_{0}

. Random jumps (about every

1 0^{- 10} s

for ordinary lamps) make

I

flicker between

0

and

4 I_{0}

far too fast to follow, so we see only the average

2 I_{0} = I_{1} + I_{2}

: no fringes.

For example, $y_{1} = A_{1} sin (ω t + k x)$ and $y_{2} = A_{2} sin (ω t + k x)$ have phase difference $0$ , and $y_{1} = A_{1} sin (ω t + k x + 3 0^{\circ})$ and $y_{2} = A_{2} sin (ω t + k x + 6 0^{\circ})$ have phase difference $3 0^{\circ}$ . Both pairs are coherent because the difference does not change with time.

Two independent sources are never coherent: atoms emit in short random bursts, so the phase jumps many times in the time an eye or detector needs to respond. Coherent sources are therefore made from one source, in one of two ways:

Division of wavefront

Two parts of the same wavefront are used as the two sources. Examples: Young's double slit, Lloyd's mirror, Fresnel's biprism.

Division of amplitude

One wave is split in strength by partial reflection and transmission. Examples: thin films, soap bubbles, Newton's rings.

6. Superposition of Two Sinusoidal Waves

Consider two waves of the same frequency meeting at a point:

x_{1} = a_{1} sin ω t, x_{2} = a_{2} sin (ω t + ϕ)

Their sum is again a sine wave of the same frequency, $x = A sin (ω t + ϕ_{0})$ , with

A^{2} = a_{1}^{2} + a_{2}^{2} + 2 a_{1} a_{2} cos ϕ, tan ϕ_{0} = \frac{a _{2} sin ϕ}{a _{1} + a _{2} cos ϕ}

The quickest way to see this is a phasor diagram: represent each wave by a rotating arrow of length equal to its amplitude, at an angle equal to its phase, and add the arrows like vectors (the same result as combining two SHMs).

Figure 10: Phasor addition. Draw

a_{1}

, then

a_{2}

at angle

ϕ

; the closing side is the resultant

A

(here

A = 5

ϕ_{0} = 5 3^{\circ}

), exactly the triangle law of vectors.

Exam Trick

Convert every term to the same function first: $cos ω t = sin (ω t + \frac{π}{2})$ . Then $3 sin ω t + 4 cos ω t$ is a 3-4-5 triangle: amplitude $5$ , phase $tan^{- 1} \frac{4}{3} = 5 3^{\circ}$ .

7. Interference: Path Difference, Phase Difference and Intensity

Let waves from coherent sources $S_{1}$ and $S_{2}$ reach a point P after travelling distances $x_{1}$ and $x_{2}$ :

y_{1} = A_{1} sin (ω t + k x_{1}), y_{2} = A_{2} sin (ω t + k x_{2}), k = \frac{2 π}{λ}

★ Must learnPhase difference from path difference:

Δ ϕ = k Δ x = \frac{2 π}{λ} Δ x

, where

Δ x = x_{2} - x_{1}

. A path difference of one wavelength equals a phase difference of

2 π

Using $A_{n e t}^{2} = A_{1}^{2} + A_{2}^{2} + 2 A_{1} A_{2} cos Δ ϕ$ and $I \propto A^{2}$ :

I_{n e t} = I_{1} + I_{2} + 2 I_{1} I_{2} cos Δ ϕ

Figure 11: Adding two waves point by point (curves computed from

y = a_{1} sin ω t + a_{2} sin (ω t + Δ ϕ)

). In phase the amplitudes add; in opposite phase they subtract.

7.1 Constructive and destructive interference

	Constructive (bright)	Destructive (dark)
Phase difference	$Δ ϕ = 2 nπ$	$Δ ϕ = (2 n + 1) π$
Path difference	$Δ x = nλ$	$Δ x = (2 n + 1) \frac{λ}{2}$
Amplitude	$A_{1} + A_{2}$	$∣ A_{1} - A_{2} ∣$
Intensity	$I_{ma x} = (I_{1} + I_{2})^{2}$	$I_{min} = (I_{1} - I_{2})^{2}$
If $I_{1} = I_{2} = I$	$4 I$	$0$ (perfect darkness)

Here $n = 0, 1, 2, 3, \dots$ For two waves of equal intensity $I_{0}$ the general result simplifies:

I = 2 I_{0} (1 + cos Δ ϕ) = 4 I_{0} cos^{2} \frac{Δ ϕ}{2}

Figure 12:

I = I_{1} + I_{2} + 2 I_{1} I_{2} cos Δ ϕ

for

I_{1} = 4 I_{0}

I_{2} = I_{0}

. Energy is only redistributed: the average stays

I_{1} + I_{2} = 5 I_{0}

(dashed line).

7.2 Maximum to minimum intensity ratio

\frac{I _{ma x}}{I _{min}} = \frac{( I _{1} + I _{2} ) ^{2}}{( I _{1} - I _{2} ) ^{2}} = (\frac{a _{1} + a _{2}}{a _{1} - a _{2}})^{2} = (\frac{r + 1}{r - 1})^{2}, r = \frac{a _{1}}{a _{2}} = \frac{I _{1}}{I _{2}}

Exam Trick

Square-root first, always. Intensity ratio $9 : 1$ means amplitude ratio $3 : 1$ , so $I_{ma x} : I_{min} = (3 + 1)^{2} : (3 - 1)^{2} = 4 : 1$ . Going backwards, $I_{ma x} / I_{min} = 25$ gives $\frac{r + 1}{r - 1} = 5$ , so $r = \frac{3}{2}$ and $I_{1} : I_{2} = 9 : 4$ .

7.3 Energy is conserved

Interference does not create or destroy energy. The average of $cos Δ ϕ$ over a pattern is zero, so the average intensity is $I_{1} + I_{2}$ , the same as without interference. Energy missing from the dark fringes appears in the bright ones.

JEE Advanced

Fringe visibility. The contrast of a pattern is $V = \frac{I _{ma x} - I _{min}}{I _{ma x} + I _{min}} = \frac{2 I _{1} I _{2}}{I _{1} + I _{2}}$ . It is 1 (best contrast) only when $I_{1} = I_{2}$ . For $n$ identical coherent sources in phase the amplitudes add, $A = na$ , so $I = n^{2} I_{0}$ ; for incoherent sources $I = n I_{0}$ .

Key idea

Coherent sources: add amplitudes with their phase. Incoherent sources: add intensities. Either way the average intensity is

I_{1} + I_{2}

Figure 13: Intensity at any point in four moves: coherent or not, optical path difference,

Δ ϕ = \frac{2 π}{λ} Δ x

, then the intensity formula.

Quick Recall: tap to check

Two coherent waves have intensities in the ratio

16 : 1

. Find

I_{ma x} : I_{min}

Amplitudes

4 : 1

, so

(4 + 1)^{2} : (4 - 1)^{2} = 25 : 9

What phase difference does a path difference of

\frac{λ}{3}

produce?

Δ ϕ = \frac{2 π}{λ} \cdot \frac{λ}{3} = \frac{2 π}{3}

Two equal coherent sources (

I_{0}

each) meet with

Δ ϕ = \frac{2 π}{3}

. Find

I

4 I_{0} cos^{2} 6 0^{\circ} = I_{0}

What is the resultant of

3 sin ω t

and

4 cos ω t

5 sin (ω t + 5 3^{\circ})

: a 3-4-5 phasor triangle.

Figure 14: Revision map: wavefronts, Huygens' principle, refraction, coherence, phasor addition and interference, with

I = I_{1} + I_{2} + 2 I_{1} I_{2} cos Δ ϕ

at the core.

8. Solved Examples

Solved Example 1

i_{1} = 3 sin ω t

and

i_{2} = 4 cos ω t

, find

i_{3} = i_{1} + i_{2}

Solution:

Write both as sines: $i_{2} = 4 sin (ω t + \frac{π}{2})$ . So $a_{1} = 3$ , $a_{2} = 4$ , $ϕ = 9 0^{\circ}$ .

$A = 3^{2} + 4^{2} + 2 (3) (4) cos 9 0^{\circ} = 5$ and $tan ϕ_{0} = \frac{4 sin 9 0 ^{\circ}}{3 + 4 cos 9 0 ^{\circ}} = \frac{4}{3}$ .

Answer: $i_{3} = 5 sin (ω t + tan^{- 1} \frac{4}{3}) = 5 sin (ω t + 5 3^{\circ})$ (Figure 10).

Solved Example 2

S_{1}

and

S_{2}

are two sources of light which individually produce disturbances at point P given by

E_{1} = 3 sin ω t

and

E_{2} = 4 cos ω t

. Assuming

E_{1}

and

E_{2}

to be along the same line, find the result of their superposition.

Solution:

Because the fields are along the same line, superposition is ordinary addition: $E = E_{1} + E_{2} = 3 sin ω t + 4 sin (ω t + \frac{π}{2})$ .

This is the same phasor triangle as Example 1: amplitude $5$ , phase lead $tan^{- 1} \frac{4}{3}$ .

Answer: $E = 5 sin (ω t + tan^{- 1} \frac{4}{3})$ , an oscillation of amplitude $5$ (same units as $E_{1}$ ) leading $E_{1}$ by $5 3^{\circ}$ .

Solved Example 3

Light from two sources, each of the same frequency and travelling in the same direction, but with intensities in the ratio

4 : 1

, interfere. Find the ratio of maximum to minimum intensity.

Solution:

$\frac{I _{1}}{I _{2}} = 4 \Rightarrow r = \frac{I _{1}}{I _{2}} = 2$ .

\frac{I _{ma x}}{I _{min}} = (\frac{r + 1}{r - 1})^{2} = (\frac{2 + 1}{2 - 1})^{2} = 9

Answer: $I_{ma x} : I_{min} = 9 : 1$ (Figure 12 shows this case).

Solved Example 4

Find the maximum intensity when

n

identical waves, each of intensity

I_{0}

, interfere if the sources are (a) coherent and (b) incoherent.

Solution:

(a) Coherent: the phase difference is constant, and the intensity is largest when all waves are in phase ( $Δ ϕ = 2 nπ$ ). Then amplitudes add. For two waves $I_{ma x} = (I_{1} + I_{2})^{2}$ ; for $n$ waves

I_{ma x} = (I_{0} + I_{0} + \dots (n terms))^{2} = n^{2} I_{0}

(b) Incoherent: the phase difference changes randomly, so $⟨ cos Δ ϕ ⟩ = 0$ and intensities simply add: $I = I_{0} + I_{0} + \dots = n I_{0}$ .

Answer: (a) $n^{2} I_{0}$ ; (b) $n I_{0}$ .

Solved Example 5

Rays of light diverge from a point source S. Draw the wavefronts. What happens to them far away from S, or after the rays pass through a convex lens with S at its focus?

Solution:

Wavefronts are always perpendicular to the rays. Rays spreading out from one point meet at right angles only circles (spheres) centred on S, so the wavefronts are spherical (Figure 3, left). A point source therefore behaves as the centre of spherical wavefronts.

Far from S only a small part of each sphere is seen, which is almost flat, and the rays are almost parallel: the wavefront becomes plane. A convex lens with S at its focus makes the rays exactly parallel, so the emerging wavefront is plane (the reverse of Figure 8, middle).

Answer: spherical wavefronts centred at S; plane wavefronts for parallel rays.

Solved Example 6

Light of wavelength

600 nm

in air enters glass of refractive index

1.5

. Find its speed, wavelength and frequency in glass. (

c = 3 \times 1 0^{8} m s^{- 1}

)

Solution:

Given: $λ = 600 nm$ , $n = 1.5$ .

Speed: $v = \frac{c}{n} = \frac{3 \times 1 0 ^{8}}{1.5} = 2 \times 1 0^{8} m s^{- 1}$ .

Wavelength: $λ_{g} = \frac{λ}{n} = \frac{600}{1.5} = 400 nm$ .

Frequency (unchanged): $ν = \frac{c}{λ} = \frac{3 \times 1 0 ^{8}}{600 \times 1 0 ^{- 9}} = 5 \times 1 0^{14} Hz$ .

Answer: $2 \times 1 0^{8} m s^{- 1}$ , $400 nm$ , $5 \times 1 0^{14} Hz$ . The light still looks orange: colour depends on frequency.

Solved Example 7

A plane wavefront in air meets a glass surface (

n = 3

) at

6 0^{\circ}

. Using Huygens' construction, find the angle the refracted wavefront makes with the surface and the ratio

A E : B C

Solution:

The angle between a wavefront and the surface equals the angle between the ray and the normal, so we need $r$ .

$sin r = \frac{sin 6 0 ^{\circ}}{3} = \frac{3 /2}{3} = \frac{1}{2} \Rightarrow r = 3 0^{\circ}$ .

$\frac{A E}{B C} = \frac{v _{2} t}{v _{1} t} = \frac{1}{n} = \frac{1}{3}$ .

Answer: the refracted wavefront makes $3 0^{\circ}$ with the surface; $A E : B C = 1 : 3$ .

Solved Example 8

Two coherent sources of equal intensity

I_{0}

produce waves of wavelength

λ

. Find the intensity at a point where the path difference is

\frac{λ}{6}

Solution:

$Δ ϕ = \frac{2 π}{λ} \cdot \frac{λ}{6} = \frac{π}{3}$ .

$I = 4 I_{0} cos^{2} \frac{Δ ϕ}{2} = 4 I_{0} cos^{2} 3 0^{\circ} = 4 I_{0} \times \frac{3}{4} = 3 I_{0}$ .

Answer: $3 I_{0}$ , which is $\frac{3}{4}$ of the maximum $4 I_{0}$ .

Solved Example 9

In an interference pattern

\frac{I _{ma x}}{I _{min}} = 25

. The ratio of the amplitudes of the interfering waves is
(A)

5 : 1

(B)

3 : 2

(C)

25 : 1

(D)

9 : 4

Solution:

$\frac{a _{1} + a _{2}}{a _{1} - a _{2}} = 25 = 5 \Rightarrow a_{1} + a_{2} = 5 a_{1} - 5 a_{2} \Rightarrow 6 a_{2} = 4 a_{1} \Rightarrow \frac{a _{1}}{a _{2}} = \frac{3}{2}$ .

Answer: (B). Amplitudes are $3 : 2$ ; option (D) $9 : 4$ is the intensity ratio, a common trap.

Solved Example 10

Two coherent waves of wavelength

500 nm

reach a point with a path difference of

1.5 μ m

. Is the point bright or dark? What if the path difference is

1.25 μ m

Solution:

$\frac{Δ x}{λ} = \frac{1500 nm}{500 nm} = 3$ , a whole number, so $Δ ϕ = 6 π$ : bright (constructive).

$\frac{1250}{500} = 2.5 = \frac{5}{2}$ , an odd multiple of $\frac{1}{2}$ , so $Δ ϕ = 5 π$ : dark (destructive).

Answer: bright for $1.5 μ m$ ; dark for $1.25 μ m$ .

Solved Example 11

The electric field of a light wave in vacuum has amplitude

E_{0} = 30 V m^{- 1}

. The wave travels along

+ x

with

E

along

y

. Find the amplitude and direction of its magnetic field.

Solution:

Given: $E_{0} = 30 V m^{- 1}$ , $c = 3 \times 1 0^{8} m s^{- 1}$ .

Amplitude: $B_{0} = \frac{E _{0}}{c} = \frac{30}{3 \times 1 0 ^{8}} = 1 \times 1 0^{- 7} T$ .

Direction: $E \times B$ points along the direction of travel. With $E$ along $\hat{j}$ and travel along $\hat{i}$ , $\hat{j} \times \hat{k} = \hat{i}$ , so $B$ is along $z$ (Figure 2).

Answer: $B_{0} = 1 \times 1 0^{- 7} T$ , along the $z$ -axis, oscillating in phase with $E$ .

Solved Example 12

Yellow light passes from air into water. Which quantity stays the same?
(A) speed
(B) wavelength
(C) frequency
(D) both speed and wavelength

Solution:

At the boundary the crests of the two sides must match, so the number of waves arriving per second equals the number leaving. Speed and wavelength both fall by the factor $n$ (Figure 6).

Answer: (C). The frequency (and hence the colour and photon energy $h ν$ ) is unchanged.

Practice Questions

Two waves of amplitudes $2 a$ and $a$ with phase difference $\frac{2 π}{3}$ superpose. Find the resultant amplitude.Answer: $A^{2} = 4 a^{2} + a^{2} + 2 (2 a) (a) (- \frac{1}{2}) = 3 a^{2}$ , so $A = 3 a$ .
Two coherent waves have intensities in the ratio $9 : 1$ . Find $I_{ma x} : I_{min}$ .Answer: $(3 + 1)^{2} : (3 - 1)^{2} = 4 : 1$ .
Sodium light ( $589 nm$ in air) enters water ( $n = \frac{4}{3}$ ). Find its wavelength and frequency in water.Answer: $λ_{w} = 442 nm$ (441.75), $ν = 5.09 \times 1 0^{14} Hz$ (same as in air).
Name the wavefront from (i) a distant star, (ii) a long tube light seen from nearby, (iii) a small bulb in a dark room.Answer: (i) plane, (ii) cylindrical, (iii) spherical.
Two identical coherent sources each give intensity $I_{0}$ . Find the intensity where their phase difference is $\frac{π}{2}$ .Answer: $4 I_{0} cos^{2} 4 5^{\circ} = 2 I_{0}$ .
A wave passes from a medium where $v_{1} = 3 \times 1 0^{8} m s^{- 1}$ into one where $v_{2} = 2.25 \times 1 0^{8} m s^{- 1}$ at $i = 5 3^{\circ}$ . Find $r$ .Answer: $sin r = 0.8 \times 0.75 = 0.6$ , so $r = 3 7^{\circ}$ .
Two independent bulbs of intensities $I$ and $4 I$ light a wall. Find the intensity on the wall.Answer: Incoherent, so $I + 4 I = 5 I$ everywhere (no fringes).

Common Mistakes to Avoid

Watch out

Thinking the frequency changes on refraction. Only speed and wavelength change: $λ_{n} = \frac{λ}{n}$ .
Adding intensities for coherent sources (use $I_{1} + I_{2} + 2 I_{1} I_{2} cos Δ ϕ$ ) or adding amplitudes for incoherent ones (use $I_{1} + I_{2}$ ).
Writing $\frac{I _{ma x}}{I _{min}} = \frac{I _{1}}{I _{2}}$ , or mixing up amplitude and intensity ratios in MCQ options ( $3 : 2$ amplitudes is $9 : 4$ intensities). Take square roots first: $(\frac{I _{1} + I _{2}}{I _{1} - I _{2}})^{2}$ .
Adding $sin$ and $cos$ terms without first converting: $cos ω t = sin (ω t + 9 0^{\circ})$ is a $9 0^{\circ}$ phase lead.
Using degrees inside $cos$ in one step and radians in the next; $Δ ϕ = \frac{2 π}{λ} Δ x$ gives radians.
Drawing rays parallel to wavefronts. Rays are always perpendicular to wavefronts.
Saying energy is destroyed at dark fringes. It is only redistributed; the average intensity is unchanged.
Calling two separate bulbs, or two halves of one lamp, coherent sources. Coherent sources must come from one source.

Frequently Asked Questions

What is a wavefront in wave optics?

A wavefront is the surface joining all points that vibrate in the same phase at an instant. Rays are always perpendicular to it. A point source gives spherical wavefronts, a line source cylindrical ones and a very distant source plane wavefronts. The gap between wavefronts differing in phase by $2 π$ is one wavelength.

What does Huygens' principle state?

Every point on a wavefront acts as a source of secondary wavelets that spread with the speed of light in that medium. After time t each wavelet has radius vt, and the forward common tangent of all the wavelets gives the new wavefront. The backward envelope is ignored because no backward wave is observed.

How does Huygens' principle prove Snell's law?

While one end of an incident wavefront travels $B C = v_{1} t$ in the first medium, the wavelet from the other end grows to $A E = v_{2} t$ in the second. Since $B C = A C sin i$ and $A E = A C sin r$ , equating the times gives $sin i / sin r = v_{1} / v_{2} = n_{21}$ , which is Snell's law.

Why can two independent bulbs not produce interference?

Light from an ordinary source comes from atoms emitting in short, random bursts, so the phase of each bulb jumps randomly many times per microsecond. The phase difference between two bulbs is not constant, the interference term averages to zero and we see only uniform brightness equal to the sum of the intensities.

Does the frequency of light change on refraction?

No. Frequency is set by the source and wavefronts must match at the boundary, so the same number of waves per second enter and leave. The speed falls to $c / n$ and the wavelength to $λ / n$ in a medium of index $n$ . Colour depends on frequency, so light keeps its colour in water or glass.

What is the ratio of maximum to minimum intensity in interference?

For waves of intensities $I_{1}$ and $I_{2}$ , $I_{ma x} / I_{min} = (I_{1} + I_{2})^{2} / (I_{1} - I_{2})^{2}$ , which equals $((a_{1} + a_{2}) / (a_{1} - a_{2}))^{2}$ in terms of amplitudes. For intensities in the ratio $4 : 1$ the amplitudes are $2 : 1$ and the ratio is $9 : 1$ .

Which wave optics basics are asked in NEET?

NEET regularly asks for the wavefront shape of a source, Huygens' principle, the fact that frequency is unchanged on refraction, the intensity formula $I = 4 I_{0} cos^{2} (Δ ϕ /2)$ and the $I_{ma x} / I_{min}$ ratio from an intensity or amplitude ratio. These are one-step questions if the square-root rule is remembered.

How is interference theory tested in JEE Main?

JEE Main uses these basics inside YDSE and thin-film problems: converting path difference to phase difference, finding intensity at a point, the ratio of maximum to minimum intensity, coherent versus incoherent addition, and phasor addition of waves such as $3 sin ω t + 4 cos ω t$ . Expect one or two questions per paper from wave optics.

Previous year questions on Introduction

2 questions from past papers, each with a step-by-step solution.

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