Back to Course
Interference Phenomena
Wavefronts and Rays
A wavefront is a surface on which all points have the same phase of vibration. A ray is a line drawn perpendicular to the wavefront, showing the direction of wave propagation.
Spherical Wavefront: Produced by a point source — concentric spheres of increasing radii, separated by one wavelength.
Plane Wavefront: At large distances from the source, a small region of a spherical wavefront appears flat. Sunlight reaching Earth has plane wavefronts.
Producing Plane Waves: Place a point source at the focus of a convex lens; the emerging rays form plane wavefronts.
Huygens' principle allows us to predict the new position of a wavefront from its current position after a time interval $\Delta t$.
Secondary Wavelets: Every point on an existing wavefront acts as a source of secondary wavelets that spread forward at speed $c$.
Envelope Construction: The new wavefront is the tangent surface (envelope) touching all secondary wavelets of radius $c\Delta t$.
Application: This principle explains both interference and diffraction by treating each slit point as a new source.
Coherence and Conditions for Interference
Coherent sources maintain a constant phase difference, which is mandatory for producing a stable, stationary interference pattern.
Monochromaticity: Light must consist of a single wavelength to avoid overlapping different patterns.
Method: Coherence is typically achieved by splitting a single wavefront into two (e.g., passing light through two narrow slits).
Why Two Bulbs Fail: Independent sources emit light in random bursts; the phase difference changes ~10⁸ times per second, washing out any pattern.
Huygens' Link: Points on the same wavefront passing through two slits are coherent secondary sources.
Conditions for Constructive and Destructive Interference
The nature of interference at a point depends on the path difference between waves arriving from the two sources.
$$d\sin\theta = m\lambda$$
Path difference condition for bright fringes (maxima) at angle θ from the central axis.
$d$=Slit separation(m)
$\theta$=Angle from central axis to the point(radians)
$m$=Order of the fringe(integer (0, ±1, ±2...))
$\lambda$=Wavelength(m)
$m = 0$
→Central bright fringe — zero path difference, always bright regardless of wavelength.
Constructive: Path difference = $0, \lambda, 2\lambda...$ → peaks align with peaks → bright maxima.
Destructive: Path difference = $0.5\lambda, 1.5\lambda...$ → peaks align with troughs → dark minima. Condition: $d\sin\theta = (m + \frac{1}{2})\lambda$.
Phase–Path Link: A path difference of $\lambda$ corresponds to a phase difference of $2\pi$ rad. General relation: $\phi = \frac{2\pi}{\lambda}\Delta L$.
Young's Double Slit Experiment (YDSE)
The spacing between adjacent bright or dark bands, known as fringe width, depends on geometry and wavelength. This is derived using the small-angle approximation $\sin\theta \approx \tan\theta = y/L$.
$$\Delta y = \frac{\lambda L}{d}$$
Fringe spacing — the distance between consecutive bright (or dark) fringes on the screen.
$\Delta y$=Fringe width (spacing between adjacent fringes)(m)
$\lambda$=Wavelength of light(m)
$L$=Distance from slits to screen(m)
$d$=Distance between slits(m)
$L \gg d$
→Small angle approximation holds: $\sin\theta \approx \tan\theta \approx y/L$.
Apparatus immersed in medium of refractive index $n$
→$\lambda' = \lambda/n$, so fringe width shrinks by factor $n$: $\Delta y' = \Delta y / n$.
Proportionality Shortcut: $\Delta y \propto \lambda L / d$ — doubling the screen distance doubles fringe width; doubling slit separation halves it.
Equal Spacing: Bright and dark fringes have the same width and are equally spaced.
Fringe Position: The $m^{th}$ bright fringe is at $y_m = m\lambda L / d$; the $m^{th}$ dark fringe is at $y_m = (m + \frac{1}{2})\lambda L / d$.
When two coherent waves of different amplitudes interfere, the resultant intensity depends on the amplitude ratio.
$$\frac{I_{max}}{I_{min}} = \left(\frac{A_1 + A_2}{A_1 - A_2}\right)^2$$
Ratio of maximum to minimum intensity in an interference pattern from sources of unequal amplitude.
$I_{max}$=Maximum intensity (constructive)(W/m²)
$I_{min}$=Minimum intensity (destructive)(W/m²)
$A_1, A_2$=Amplitudes of the two waves(m)
$A_1 = A_2$
→$I_{min} = 0$ — complete destructive interference. Dark fringes are truly dark.
Intensity–Amplitude Link: $I \propto A^2$, so if intensity ratio is given, take the square root to get the amplitude ratio first.
Resultant Amplitude: $A_{res} = \sqrt{A_1^2 + A_2^2 + 2A_1 A_2 \cos\phi}$, where $\phi$ is the phase difference.
Limiting Case: When $\phi = 0$ (constructive), $A_{res} = A_1 + A_2$; when $\phi = \pi$ (destructive), $A_{res} = |A_1 - A_2|$.
Interference in Thin Films
A thin film is a transparent layer whose thickness is comparable to the wavelength of light. Colours in soap bubbles and oil on water arise from interference between light reflected from the film's two surfaces.
How It Works: An incident ray partly reflects from the top surface and partly refracts into the film. The refracted ray reflects off the bottom surface and emerges parallel to the first reflected ray. These two coherent reflected rays interfere.
Path Difference Depends On: (i) thickness of the film, (ii) refractive index of the film, and (iii) angle of incidence.
Phase Reversal on Reflection: When light reflects from a denser medium (low-n to high-n), it undergoes a phase shift of $\pi$ (equivalent to an extra $\lambda/2$ path difference). This is critical for determining constructive vs destructive conditions.
White Light Colours: Different thicknesses satisfy destructive interference for different colours. The remaining colours are seen — this is why oil films show rainbow patterns.
Dimensional Insight: The film thickness must be on the order of wavelengths (~400–700 nm) for visible interference effects.
Newton's Rings
When a plano-convex lens of long focal length is placed on a flat glass plate, the air gap between them forms a thin film of varying thickness, producing concentric circular interference fringes called Newton's rings.
Air Wedge: The air film thickness is zero at the contact point and increases outward. Points of equal thickness lie on circles centred at the contact point.
Dark Centre: At the contact point the film thickness is effectively zero, but reflection from the denser glass plate introduces an extra $\lambda/2$ path difference, causing destructive interference — the centre is always dark.
Ring Pattern: Alternating bright and dark rings appear. The rings get closer together as you move outward (ring spacing decreases with increasing order).
Michelson's Interferometer
The Michelson interferometer splits a beam into two perpendicular paths using a half-silvered mirror, reflects them back, and recombines them to produce interference fringes.
$$L = m\frac{\lambda}{2}$$
Relates the mirror displacement to the number of fringes shifted.
$L$=Displacement of the movable mirror(m)
$m$=Number of fringes shifted past the reference(integer)
$\lambda$=Wavelength of light used(m)
$m = 1$
→One fringe shift corresponds to a mirror displacement of $\lambda/2$.
Beam Splitter: A half-silvered plate ($G_1$) divides the incoming beam into two perpendicular paths.
Compensator Plate: A second plate ($G_2$) equalises the optical path through glass for both beams.
Factor of 2: Moving the mirror by $\lambda/2$ changes the round-trip path by $\lambda$, shifting one full fringe.
Precision: A displacement of $\lambda/4$ produces a visible change from bright to dark. For $\lambda = 400$ nm, this means ~100 nm precision ($10^{-7}$ m).
Application Strategy
Worked Examples
Problem Statement
The distance between the slits in YDSE is unknown. The screen is 100 cm away and the 3rd dark fringe is 0.059 cm from centre. If $\lambda = 590$ nm, verify the slit separation.
Dark fringe position formulaUnit conversion
Strategic Toolkit
$$y = \left(m + \frac{1}{2}\right)\frac{\lambda L}{d}$$
Step 1
Identify Order
The 3rd dark fringe corresponds to $m = 2$ (since the 1st dark has $m = 0$).
$m = 2$
Step 2
Convert Units
Express everything in SI: $y = 5.9 \times 10^{-4}$ m, $L = 1$ m, $\lambda = 5.9 \times 10^{-7}$ m.
$y = 5.9 \times 10^{-4}$ m
Step 3
Rearrange and Solve
Solve for $d$: $d = (m + 0.5)\lambda L / y$.
$d = \frac{2.5 \times 5.9 \times 10^{-7} \times 1}{5.9 \times 10^{-4}} = 2.5 \times 10^{-3}$ m = 2.5 mm
Problem Statement
Yellow sodium light ($\lambda = 589$ nm) passes through two slits 1.00 mm apart. The screen is 225 cm away. Find the fringe spacing.
Fringe width formulaProportionality
Strategic Toolkit
$$\Delta y = \frac{\lambda L}{d}$$
Step 1
Convert Units
All to metres: $\lambda = 589 \times 10^{-9}$ m, $d = 1.00 \times 10^{-3}$ m, $L = 2.25$ m.
$\lambda = 5.89 \times 10^{-7}$ m
Step 2
Substitute
Apply the fringe width formula directly.
$\Delta y = \frac{5.89 \times 10^{-7} \times 2.25}{1.00 \times 10^{-3}} = 1.33 \times 10^{-3}$ m
Step 3
Result
The adjacent bright fringes are 1.33 mm apart.
$\Delta y = 1.33$ mm
Problem Statement
In a Michelson interferometer using light of wavelength 600 nm, the movable mirror is displaced and 200 fringe shifts are counted. How far did the mirror move?
Michelson formulaFactor of 2 in round-trip
Strategic Toolkit
$$L = m\frac{\lambda}{2}$$
Step 1
Identify Knowns
$m = 200$ fringes, $\lambda = 600 \times 10^{-9}$ m.
$m = 200$, $\lambda = 6 \times 10^{-7}$ m
Step 2
Substitute
Apply the Michelson formula.
$L = 200 \times \frac{6 \times 10^{-7}}{2} = 6 \times 10^{-5}$ m
Step 3
Result
The mirror moved 0.06 mm (60 μm).
$L = 60$ μm
Memorize formulas
Fringe Width (YDSE)
'Lambda-L over d' — the spread of interference bands.
Formula
$\Delta y = \frac{\lambda L}{d}$
Bright Fringe Condition
Integer wavelengths path difference → constructive.
Formula
$d\sin\theta = m\lambda$
Dark Fringe Condition
Half-integer wavelengths → destructive.
Formula
$d\sin\theta = (m + \tfrac{1}{2})\lambda$
Bright Fringe Position
Linear position of the mth bright fringe on the screen.
Formula
$y_m = \frac{m\lambda L}{d}$
Phase–Path Relation
Converts physical path difference to phase angle.
Formula
$\phi = \frac{2\pi}{\lambda}\Delta L$
Intensity Ratio
Max-to-min intensity from unequal amplitude sources.
Formula
$\frac{I_{max}}{I_{min}} = \left(\frac{A_1 + A_2}{A_1 - A_2}\right)^2$
Michelson Interferometer
Mirror displacement from fringe count.
Formula
$L = m\frac{\lambda}{2}$
Review
1
In YDSE with $d = 0.5$ mm and $L = 1.0$ m, a thin glass sheet of thickness 0.1 mm and refractive index 1.5 is placed over one slit. By how much does the central bright fringe shift?
Ans
0.1 mm
Review
Tap to reveal
Find the optical path delay
A glass sheet adds only the excess optical path relative to air, which is $(n-1)t$.
$$\Delta_{opt}=(1.5-1)(0.1\,\text{mm})=0.05\,\text{mm}$$
Convert delay to screen shift
In YDSE, a path delay shifts the central fringe by $\Delta_{opt}L/d$.
$$\Delta y=\frac{(n-1)tL}{d}=\frac{0.05\times10^{-3}\times1.0}{0.5\times10^{-3}}=0.1\,\text{mm}$$
2
Two coherent sources have amplitudes in the ratio 3:1. If the maximum intensity in the interference pattern is $I_0$, what is the minimum intensity?
Ans
$I_0/4$
Review
Tap to reveal
Write the extreme intensities from amplitudes
Let the amplitudes be $3A$ and $A$. The maximum and minimum intensities come from adding and subtracting these amplitudes.
$$I_{\max}=(3A+A)^2=16A^2,\quad I_{\min}=(3A-A)^2=4A^2$$
Compare minimum to maximum
The ratio is $4/16$, so the minimum is one-fourth of the maximum.
$$I_{\min}=\frac{1}{4}I_0$$
3
In YDSE, the phase difference between the two waves arriving at a certain point on the screen is $8\pi$ rad. This point corresponds to:
Ans
The 4th bright fringe
Review
Tap to reveal
Use the bright-fringe phase condition
Bright fringes occur when the phase difference is an integer multiple of $2\pi$.
$$8\pi=2m\pi\Rightarrow m=4$$
4
In YDSE, if the slit separation $d$ is doubled and the screen distance $L$ is halved simultaneously, the fringe width will:
Ans
Become one-fourth
Review
Tap to reveal
Apply the fringe-width formula
Fringe width is proportional to $L/d$, so doubling $d$ and halving $L$ both reduce it.
$$\beta' = \frac{\lambda (L/2)}{2d} = \frac{1}{4}\frac{\lambda L}{d} = \frac{\beta}{4}$$
5
In Newton's rings experiment, if the pattern is observed in transmitted light instead of reflected light, the centre of the pattern will be:
Ans
Bright, because there is no net phase reversal for transmitted rays
Review
Tap to reveal
Use the transmitted-light complement
In transmitted light, the two rays do not pick up the same extra half-wavelength reversal that makes the reflected center dark.
At zero thickness, the centre is bright.
6
In a Michelson interferometer using $\lambda = 500$ nm, a thin glass plate of thickness 0.01 mm is inserted in one arm. If 40 fringe shifts are observed, what is the refractive index of the glass?
Ans
1.5
Review
Tap to reveal
Write the fringe-shift relation
The plate is traversed twice, so the extra optical path is $2(n-1)t$ and the shift count is that path divided by $\lambda$.
$$m=\frac{2(n-1)t}{\lambda}$$
Solve for the refractive index
Substitute the numbers and isolate $n$.
$$n=1+\frac{m\lambda}{2t}=1+\frac{40\times500\times10^{-9}}{2\times0.01\times10^{-3}}=1.5$$
7
Thin film interference colours are seen in oil on water because:
Ans
Light reflected from the top and bottom surfaces of the oil film interferes constructively for some wavelengths and destructively for others
Review
Tap to reveal
Identify the interfering beams
The observed colour comes from reflection at the top and bottom surfaces of the thin oil layer.
Some wavelengths reinforce, while others cancel.
8
According to Huygens' principle, each point on a wavefront acts as a source of secondary wavelets. The new wavefront is formed by:
Ans
The forward tangent envelope of all secondary wavelets
Review
Tap to reveal
Use the envelope construction
Huygens' principle forms the next wavefront from the forward common tangent to all secondary wavelets.
Forward tangent envelope of the secondary wavelets.
9
In YDSE with $d = 0.2$ mm and $L = 1.5$ m, two wavelengths $\lambda_1 = 600$ nm and $\lambda_2 = 450$ nm are used simultaneously. At what minimum distance from the central maximum do bright fringes of both wavelengths coincide?
Ans
13.5 mm
Review
Tap to reveal
Compute both fringe widths
Use $\beta=\lambda L/d$ for each wavelength.
$$\beta_1=4.5\,\text{mm},\quad \beta_2=3.375\,\text{mm}$$
Match the fringe positions
Coincidence requires $m_1\beta_1=m_2\beta_2$. The smallest integer ratio is $3:4$.
$$y=3\beta_1=3\times4.5=13.5\,\text{mm}$$
10
Two coherent waves of equal amplitude $A$ meet at a point where their phase difference is $2\pi/3$. What is the resultant amplitude?
Ans
$A$
Review
Tap to reveal
Use the phasor amplitude formula
For two equal amplitudes, the resultant amplitude is $2A\cos(\phi/2)$.
$$A_{res}=2A\cos\left(\frac{2\pi/3}{2}\right)=2A\cos\left(\frac{\pi}{3}\right)=A$$
Question 1 of 10
0:00
1
In YDSE with $d = 0.5$ mm and $L = 1.0$ m, a thin glass sheet of thickness 0.1 mm and refractive index 1.5 is placed over one slit. By how much does the central bright fringe shift?