8A · Interference — YDSE, Coherence, Energy, Fringe Shift

Coherence conditions, why two sources fail, full YDSE intensity derivation, fringe width and equal-width proof, energy conservation, two solved numerics.

YDSE means Young’s double-slit experiment. One lamp lights two narrow slits. The two beams overlap on a screen and give equal bright and dark stripes. Daily example: two water ripples crossing in a pond.

Formula 1. Resultant intensity in YDSE — $I = I_1 + I_2 + 2\sqrt{I_1 I_2}\cos\delta$

Formula 2. Fringe width and positions — $y_n = \dfrac{n\lambda D}{d}$ (bright), $\beta = \dfrac{\lambda D}{d}$

Here $I_1$, $I_2$ = intensities from slit 1 and slit 2. $\delta$ = phase difference between the two beams. $\lambda$ = wavelength of light. $d$ = slit separation. $D$ = slit-to-screen distance. $y_n$ = position of the $n$-th bright fringe from the centre. $\beta$ = fringe width (one full stripe spacing).

Young double slit geometry with slit separation d, screen distance D, fringe width beta

Fig: YDSE — two slits ($d$ apart) send waves to a screen ($D$ away); path difference $d\sin\theta$ gives $I(\theta)$ in Eq. (1).

How YDSE works (short derivation)

Step 1 — Coherent waves. At a screen point, the two displacements are $y_1 = a_1\sin\omega t$ and $y_2 = a_2\sin(\omega t + \delta)$. Same $\omega$ and fixed $\delta$ make them coherent.

Step 2 — Add them. By superposition, $y = y_1 + y_2 = A\sin(\omega t + \phi)$ with $A^2 = a_1^2 + a_2^2 + 2a_1 a_2\cos\delta$.

Step 3 — Amplitude → intensity. Light intensity $I \propto A^2$, so with $I_1 \propto a_1^2$, $I_2 \propto a_2^2$ we get Eq. (1): $I = I_1 + I_2 + 2\sqrt{I_1 I_2}\cos\delta$. The $\cos\delta$ term is the interference term — it is missing for incoherent light.

Step 4 — Phase from path. Path difference $\Delta = S_2P - S_1P = d\sin\theta \approx dy/D$ for small angles. Phase $\delta = (2\pi/\lambda)\Delta$.

Step 5 — Bright/dark. Bright (max): $\cos\delta = +1$, so $\delta = 2n\pi$, $\Delta = n\lambda$, positions $y_n = n\lambda D/d$. Dark (min): $\cos\delta = -1$, so $\delta = (2n+1)\pi$, $\Delta = (2n+1)\lambda/2$, positions $y_n’ = (2n+1)\lambda D/2d$.

Step 6 — Fringe width. $\beta = y_{n+1} - y_n = \lambda D/d$. Dark spacing is the same: $y_{n+1}’ - y_n’ = \lambda D/d$. So bright and dark bands have equal width.

Unit check: $[\lambda D/d] = (L \cdot L)/L = L$, correct for $\beta$. Limit: $d \to 0$ gives $\beta \to \infty$ (slits merge, no stripes — sensible).

Plot: $I$ vs $y$ is a $\cos^2$ curve — peaks $I_{max} = I_1+I_2+2\sqrt{I_1I_2}$ (equals $4I_0$ if $I_1 = I_2 = I_0$), valleys $I_{min} = I_1+I_2-2\sqrt{I_1I_2}$ (zero if equal). Draw equal-height peaks, equal spacing $\beta$.

Trap: $\beta$ uses $d$ (slit gap), not slit width. Do not mix them.

Solved PYQs

What is interference of light? State the fundamental conditions for a steady observable interference pattern.

2019

Recall

Interference = redistribution of intensity by superposition of two coherent waves. Without coherence the $\cos\delta$ term averages to zero.

Steps

Define interference: when two coherent waves overlap, intensity varies point to point — bright and dark fringes appear.
Condition (i) — coherent sources: same frequency and a constant phase difference, else $\delta$ jumps $\sim 10^8$ times per second and the eye sees only the average $I_1+I_2$.
Condition (ii) — comparable intensities, else minima stay bright ($I_{min} = (\sqrt{I_1}-\sqrt{I_2})^2$ large) and contrast is poor.
Condition (iii) — same state of polarization, because perpendicular vibrations do not interfere and intensities just add.
Condition (iv) — narrow slits close together and (ideally) monochromatic light, because wide or polychromatic sources overlap many patterns and wash fringes out.

Answer

Steady fringes need coherence + comparable amplitudes + same polarization + narrow slits (monochromatic for sharp fringes). $\boxed{\text{4 conditions: coherence, equal amplitude, same polarization, narrow slits (monochromatic)}}$

Exam tip

Write the 4 conditions as 4 numbered lines — examiners give one mark per condition.

"Two separate sources of light cannot produce interference of light" — explain.

2022

Recall

Atoms emit in short bursts ($\sim 10^{-8}$ s). Each burst has a random starting phase, so two independent lamps have a random phase difference.

Steps

Two independent lamps = billions of independent atoms, each emitting in bursts of $\sim 10^{-8}$ s with random starting phase.
So the phase difference $\delta(t)$ between the lamps changes randomly $\sim 10^8$ times per second — not coherent.
Eye or screen averages over milliseconds, so $\langle\cos\delta\rangle = 0$ and observed $I = I_1 + I_2$ — uniform light, no fringes.
Cure: take ONE source and split it — division of wavefront (YDSE, biprism) or division of amplitude (thin films, Newton rings) — so both beams share the same random jumps and $\delta$ stays fixed.

Answer

$\boxed{\text{Two separate sources are incoherent; split one source in two to get steady fringes.}}$

Exam tip

Draw the contrast: independent sources (random $\delta$) vs split source (fixed $\delta$). One diagram earns full marks.

Young's double slit experiment: coherent waves, resultant intensity pattern, conditions of maxima and minima (both phase and path), and the intensity distribution plot.

2019

Recall

Eqs. (1)–(2): $I = I_1+I_2+2\sqrt{I_1I_2}\cos\delta$ and $\beta = \lambda D/d$. Bright when $\cos\delta = +1$, dark when $\cos\delta = -1$.

Steps

Take coherent waves $a_1\sin\omega t$ and $a_2\sin(\omega t+\delta)$ — coherence assumed else there is no steady $\delta$.
Derive $I = I_1+I_2+2\sqrt{I_1I_2}\cos\delta$ using superposition and $I \propto A^2$.
Maxima: $\cos\delta = +1$, so $\delta = 2n\pi$ (phase) and $\Delta = n\lambda$ (path); $I_{max} = (\sqrt{I_1}+\sqrt{I_2})^2$.
Minima: $\cos\delta = -1$, so $\delta = (2n+1)\pi$ (phase) and $\Delta = (2n+1)\lambda/2$ (path); $I_{min} = (\sqrt{I_1}-\sqrt{I_2})^2$.
Plot $I$ vs $y$: a $\cos^2$ curve with equal peaks spaced $\beta = \lambda D/d$, touching zero if $I_1 = I_2$.

Answer

$\boxed{I = I_1+I_2+2\sqrt{I_1I_2}\cos\delta;\quad \text{bright }\Delta = n\lambda;\quad \text{dark }\Delta = (2n+1)\lambda/2}$

Exam tip

Always give BOTH the phase form ($\delta$) and the path form ($\Delta$) — the paper asks for both and missing one loses a mark.

State the conditions for a steady interference pattern. Find the fringe width in YDSE and prove that the dark and bright bands are of equal width.

2024

Recall

Steady-pattern conditions from Q1 + bright positions $y_n = n\lambda D/d$ and dark positions $y_n' = (2n+1)\lambda D/2d$.

Steps

State the 4 steady-pattern conditions briefly: coherence, comparable amplitudes, same polarization, narrow slits (monochromatic).
Path difference $\Delta = dy/D$ and phase $\delta = 2\pi\Delta/\lambda$ from small-angle geometry.
Bright positions: $y_n = n\lambda D/d$, so $\beta_{bright} = y_{n+1}-y_n = \lambda D/d$.
Dark positions: $y_n' = (2n+1)\lambda D/2d$, so $\beta_{dark} = y_{n+1}'-y_n' = \lambda D/d$.
Hence $\beta_{bright} = \beta_{dark} = \lambda D/d$ — equal width proved.

Answer

$\boxed{\beta = \dfrac{\lambda D}{d}\ \text{for bright and dark alike.}}$

Exam tip

The proof is two explicit subtractions — write both out, never say "similarly" without showing.

What are constructive and destructive interference? Does destructive interference violate the law of conservation of energy?

20222023×2

Recall

Constructive = in phase ($\delta = 2n\pi$); destructive = out of phase ($\delta = (2n+1)\pi$). The average $\langle\cos\delta\rangle = 0$ over one fringe.

Steps

Constructive: waves in step, amplitudes add — $I_{max} = I_1+I_2+2\sqrt{I_1I_2}$ (equals $4I_0$ if $I_1=I_2=I_0$).
Destructive: waves opposite, amplitudes subtract — $I_{min} = I_1+I_2-2\sqrt{I_1I_2}$ (zero if equal).
Energy is not destroyed: average intensity over one fringe pair = $I_1 + I_2$ (the $\cos\delta$ term averages to zero).
So energy is REDISTRIBUTED — taken from dark places and piled onto bright places; total stays $I_1+I_2$ times area.

Answer

No violation. Energy is redistributed, not destroyed. $\boxed{\langle I\rangle = I_1+I_2}$ — over a full fringe, dark loses and bright gains; total is the same.

Exam tip

This is a repeat (×2). Memorise one line: "dark loses, bright gains, total same."

In YDSE the slit separation $d = 0.1$ mm, the fringe width $\beta = 5$ mm, and the screen distance $D = 1$ m. Find the wavelength $\lambda$.

2024

Recall

From $y_n = n\lambda D/d$ the spacing is $\beta = \lambda D/d$, so $\lambda = \beta d/D$.

Steps

Convert to metres: $d = 0.1\ \text{mm} = 1.0\times 10^{-4}$ m; $\beta = 5\ \text{mm} = 5.0\times 10^{-3}$ m; $D = 1$ m.
Formula: from $\beta = \lambda D/d$, rearrange to $\lambda = \beta d/D$.
Substitute: $\lambda = (5.0\times 10^{-3})(1.0\times 10^{-4})/1 = 5.0\times 10^{-7}$ m.
Convert for sense: $5.0\times 10^{-7}$ m $= 500$ nm — green light, inside the visible band (400–700 nm).

Answer

$\boxed{\lambda = 5\times 10^{-7}\ \text{m} = 500\ \text{nm}}$

Exam tip

Always convert mm to m before multiplying — most errors come from mixing mm and m in one line.

In YDSE a glass plate ($\mu_g = 1.5$, thickness $t_g = 12\times 10^{-5}$ mm) is placed over one slit and a diamond plate ($\mu_d = 2.5$) over the other. The central fringe does not shift. Find the thickness of the diamond plate.

2023

Recall

A plate of index $\mu$ and thickness $t$ adds extra path $(\mu-1)t$ and shifts fringes by $(\mu-1)tD/d$ toward that side.

Steps

Glass adds extra path $(\mu_g-1)t_g$ on the slit-1 side; diamond adds $(\mu_d-1)t_d$ on the slit-2 side.
No shift of central fringe means the two extra paths cancel: $(\mu_g-1)t_g = (\mu_d-1)t_d$.
Solve for $t_d = (\mu_g-1)t_g/(\mu_d-1) = (1.5-1)(12\times 10^{-5})/(2.5-1)$.
Arithmetic: numerator $= 0.5 \times 12 \times 10^{-5} = 6 \times 10^{-5}$; divided by $1.5$ gives $t_d = 4 \times 10^{-5}$ mm.
Sense: diamond bends more ($\mu$ larger), so a thinner piece balances the glass — $4 < 12$, sensible.

Answer

$\boxed{t_d = 4\times 10^{-5}\ \text{mm}}$

Exam tip

"No shift" always means "extra paths equal". Write that line first, then substitute — that ordering is what gets the first mark.

Reference. Brij Lal, Subrahmanyam & Avadhanulu — Optics Ch-13 (coherence), Ch-14 (YDSE, fringe shift, energy); Ajoy Ghatak — Optics Ch-13.