8D · Interference — Lloyd's Mirror

Lloyd's mirror setup, virtual source by reflection, Stokes half-wave flip, fringe width, distinction from Fresnel biprism, and three solved fringe-width numerics.

A slit sits just above a flat glass mirror. Light reaches the screen two ways — straight from the slit, and bounced off the mirror. The bounced beam looks as if it comes from a mirror-image slit below the glass. The two beams interfere. Daily example: your reflection in a shop window overlapping with you.

Formula 1. Fringe width — $\beta = \dfrac{\lambda D}{d}$ (same as YDSE; $d$ = slit to virtual-image separation, $D$ = slit-plane to screen distance)

Here $S$ = real slit (height $h$ above the mirror), $S’$ = virtual image ($h$ below the mirror plane), so $d = 2h$. $D$ = distance from the source plane to the screen.

Lloyd's mirror with real source S above glass plate, virtual image S-prime below, reflected ray with half-wave flip, screen showing overlap fringes

Fig: Lloyd's mirror — $S$ interferes with its own mirror image $S'$; the reflected beam picks up a $\lambda/2$ phase flip on reflection.

How it works

Step 1 — Setup. A narrow slit $S$ is placed a small height $h$ above a polished glass plate (the mirror), with the slit parallel to the plate surface. A screen stands vertically at distance $D$, its lower edge nearly touching the mirror.

Step 2 — Two beams. Beam 1 goes directly from $S$ to the screen. Beam 2 reflects off the glass and then reaches the screen. Tracing beam 2 backward, it appears to come from $S’$, the mirror image of $S$ at depth $h$ below the plate — so $d = SS’ = 2h$.

Step 3 — Coherence. Both beams come from the same slit $S$, so the phase difference is fixed — division of wavefront, like the biprism.

Step 4 — The Stokes flip. Reflection at the glass (rarer→denser) flips the phase by $\pi$, i.e. adds $\lambda/2$ to the reflected beam’s path. So where YDSE would give a bright centre, Lloyd’s mirror gives a DARK centre — the central fringe (at the mirror plane level) is dark.

Step 5 — Fringes. Path difference $\Delta = dy/D$ as usual, but with the extra $\lambda/2$: bright where $\Delta = (n-\tfrac12)\lambda$, dark where $\Delta = n\lambda$. Fringe width is unchanged: $\beta = \lambda D/d$.

Unit check: $[\lambda D/d] = L$ ✓. Limit: $h \to 0$ ($d \to 0$) gives $\beta \to \infty$ — slit touching the mirror, no fringes, sensible.

Trap: only HALF the usual YDSE field is visible — the mirror blocks the lower half, so fewer fringes are seen than with a biprism.

Lloyd’s mirror vs Fresnel biprism (exam table)

Point	Lloyd’s mirror	Fresnel biprism
Second source	virtual image by REFLECTION ($S’$)	two virtual images by REFRACTION
Phase flip	yes — $\lambda/2$ on reflection, centre dark	no flip, centre bright
Fringes visible	half field (mirror blocks rest)	full overlap field
Extra use	proves the $\lambda/2$ Stokes flip directly	measures $d$ by displacement method

Solved PYQs

Distinguish between Lloyd's mirror and Fresnel's biprism.

2022

Recall

Mirror = reflection image + $\lambda/2$ flip + dark centre; biprism = refraction images + no flip + bright centre.

Steps

Source pair: Lloyd's uses one real slit plus its mirror image (reflection); biprism uses two virtual images by refraction through thin prisms.
Phase: the reflected beam in Lloyd's mirror suffers a $\pi$ ($\lambda/2$) flip at the glass — the biprism has no reflection, so no flip.
Centre fringe: dark for Lloyd's (flip), bright for biprism (no flip).
Field: Lloyd's shows fringes only above the mirror plane (half field); biprism shows the full overlap region.
Write the 4-row table above — the table itself is the answer.

Answer

$\boxed{\text{Mirror: reflection + }\lambda/2\text{ flip + dark centre + half field; biprism: refraction + no flip + bright centre + full field.}}$

Exam tip

The table is the answer — four rows, each row one mark. Never write it as a paragraph.

In Lloyd's mirror experiment, the real and virtual sources are $1$ mm apart and the screen is placed at a distance of $1$ m from the source. If the wavelength of light used is $5890$ Å, find the fringe width.

2022

Recall

$\beta = \lambda D/d$ — identical to YDSE; the mirror only changes how $S'$ is made.

Steps

Given: $d = 1$ mm $= 1.0\times 10^{-3}$ m; $D = 1$ m; $\lambda = 5890$ Å $= 5.89\times 10^{-7}$ m.
Formula: $\beta = \lambda D/d$.
Substitute: $\beta = (5.89\times 10^{-7}\times 1)/(1.0\times 10^{-3}) = 5.89\times 10^{-4}$ m.
Convert: $5.89\times 10^{-4}$ m $= 0.589$ mm.

Answer

$\boxed{\beta = 5.89\times 10^{-4}\ \text{m} \approx 0.59\ \text{mm}}$

Exam tip

Lloyd's numerics use the same formula as YDSE — do not invent a new one. The flip only moves bright/dark positions, not $\beta$.

In Lloyd's mirror experiment, the real and virtual sources are $1.5$ mm apart and the screen is placed at a distance of $1.5$ m from the source. If the wavelength of light used is $5890$ Å, find the fringe width.

2023

Recall

$\beta = \lambda D/d$. Note $D/d = 1.5/1.5\times 10^{-3} = 1000$ — same ratio as Q2.

Steps

Given: $d = 1.5\times 10^{-3}$ m; $D = 1.5$ m; $\lambda = 5.89\times 10^{-7}$ m.
Ratio: $D/d = 1.5/(1.5\times 10^{-3}) = 1000$.
$\beta = \lambda\times 1000 = 5.89\times 10^{-7}\times 10^3 = 5.89\times 10^{-4}$ m $= 0.589$ mm.
Sense: identical to Q2 because $D/d$ is identical — the paper scaled both numbers by $1.5\times$.

Answer

$\boxed{\beta \approx 0.59\ \text{mm}}$

Exam tip

Spot the scaling: both $D$ and $d$ grew by $1.5\times$, so $\beta$ is unchanged. One line of reasoning saves the arithmetic.

In Lloyd's mirror experiment, the real and virtual sources are $1.2$ mm apart and the screen is placed at a distance of $1.2$ m from the source. If the wavelength of light used is $5890$ Å, find the fringe width.

2024

Recall

$\beta = \lambda D/d$; again $D/d = 1000$.

Steps

Given: $d = 1.2\times 10^{-3}$ m; $D = 1.2$ m; $\lambda = 5.89\times 10^{-7}$ m.
$D/d = 1.2/(1.2\times 10^{-3}) = 1000$.
$\beta = 5.89\times 10^{-7}\times 10^3 = 5.89\times 10^{-4}$ m $\approx 0.59$ mm.

Answer

$\boxed{\beta \approx 0.59\ \text{mm}}$

Exam tip

Three years, same answer ($0.59$ mm) — the paper only rescales $D$ and $d$ together. Learn the pattern, not the numbers.

Reference. Brij Lal, Subrahmanyam & Avadhanulu — Optics Ch-14 (Lloyd's mirror, Stokes flip); Ajoy Ghatak — Optics Ch-13.