Young's double slit — fringe width derivation: Question bank

Reminder of the symbols before we abuse them:

$λ$ = wavelength (the wave's built-in ruler),
$d$ = slit separation, $D$ = slit-to-screen distance,
$y$ = height of a point on the screen above centre $O$ ,
$Δ = \frac{y d}{D}$ = path difference, $β = \frac{λ D}{d}$ = fringe width.

True or false — justify

True or false: The central fringe (at

y = 0

) is always bright.

True — at the centre both waves travel equal distances, so

Δ = 0

, which is

nλ

with

n = 0

: perfectly in phase, maximum brightness.

True or false: Fringe width is larger for the 10th fringe than for the 1st.

False — spacing between neighbours is

λ D / d

, independent of

n

; the

n

's cancel in

y_{n + 1} - y_{n}

, so all gaps are identical.

True or false: If you make the two slits closer together, the fringes spread out.

True —

β \propto 1/ d

, so smaller

d

gives a wider pattern; the sources being closer makes their overlap fan out more.

True or false: Using red light instead of blue makes fringes narrower.

False —

β \propto λ

and red has the larger wavelength, so red gives wider fringes.

True or false: Immersing the whole setup in water widens the fringes.

False — in water

λ^{'} = λ / n_{w}

shrinks, so

β^{'} = β / n_{w}

; the pattern contracts (fringes crowd closer).

True or false: Angular fringe width

λ / d

changes if you move the screen farther away.

False — angular spacing

θ = β / D = λ / d

has no

D

in it; moving the screen stretches

β

and

D

together, leaving the angle fixed.

True or false: A path difference of exactly

λ /2

gives a dark fringe.

True — half a wavelength puts a crest against a trough (out of phase), so the waves cancel; this is the

n = 0

dark condition

(n + \frac{1}{2}) λ

True or false: If the two slits emitted light of slightly different frequencies, you'd still see a stable fringe pattern.

False — different frequencies means the phase relationship keeps changing, so bright and dark spots wash out; you need coherent sources of the same frequency.

True or false: Two independent light bulbs behind two slits would produce clear YDSE fringes.

False — independent bulbs have randomly jumping phases, so any interference pattern flickers and averages to uniform brightness; that's why one source is split into two slits.

Spot the error

Error hunt: "Dark fringes occur where

Δ = nλ /2

for any integer

n

Wrong — at

n = 2

this gives

Δ = λ

, which is bright. The correct dark condition is

Δ = (n + \frac{1}{2}) λ

, i.e. odd multiples of

λ /2

Error hunt: "

Δ = d sin θ

and

Δ = y d / D

are two unrelated formulas."

Wrong — they are the same thing; since

sin θ \approx tan θ = y / D

in the Small angle approximation,

d sin θ \approx y d / D

Error hunt: "Since

S_{1} P + S_{2} P \approx 2 D

, the two path lengths are equal, so

Δ = 0

everywhere."

Wrong — the sum is approximately

2 D

, but the tiny difference

Δ = 2 y d / (S_{1} P + S_{2} P) \approx y d / D

is exactly what we keep; approximating the sum doesn't kill the difference.

Error hunt: "The formula

y_{n} = nλ D / d

works even when

y

is comparable to

D

Wrong — it relies on

D ≫ y

so that

sin θ \approx tan θ

; for large angles the small-angle approximation breaks and fringes are no longer evenly spaced.

Error hunt: "Frequency of light changes when it enters water, which is why

β

changes."

Wrong — frequency is set by the source and stays fixed; it's the speed (and hence wavelength) that drops by the refractive index

n_{w}

Error hunt: "The path difference is the height

y

of the point on the screen."

Wrong —

y

is a screen distance; the path difference is

Δ = y d / D

, which mixes in

d

and

D

to convert screen position into an extra path length.

Why questions

Why does

n

cancel when computing fringe width?

Because

y_{n} = nλ D / d

is linear in

n

, so consecutive positions differ by a constant

λ D / d

regardless of which fringe you pick — this is exactly why fringes are equally spaced.

Why do we require

D ≫ d

in the derivation?

So that both path lengths are

\approx D

and

S_{1} P + S_{2} P \approx 2 D

; this is what turns the exact geometry into the clean result

Δ = y d / D

Why must the two sources be coherent to see fringes?

A steady pattern needs a constant phase relationship between the waves; incoherent sources have phases that jump randomly, smearing bright and dark into uniform grey — see Coherence and coherent sources.

Why does bigger

λ

give wider fringes?

A larger wavelength means you need to move farther along the screen before

Δ

builds up to the next whole

λ

, so consecutive bright spots sit farther apart.

Why does the pattern shrink in water rather than stay the same?

In water the wavelength shrinks to

λ / n_{w}

, and since

β \propto λ

, the fringe spacing shrinks by the same factor

n_{w}

— see Refractive index.

Why do we split ONE source into two slits instead of using two separate lamps?

Splitting one wavefront guarantees the two slit-waves share the same phase history (they're coherent), which two independent lamps can never provide.

Why is the angular fringe width

λ / d

often more useful than

β

It drops out

D

, so it describes the intrinsic geometry of the two-source system independent of how far away you place the screen.

Edge cases

Edge case: What happens to the fringe pattern if

d \to 0

(slits merge)?

β = λ D / d \to \infty

— the fringes spread infinitely wide, meaning a single slit gives no two-source interference at all.

Edge case: What if

d

becomes very large compared to

λ

β \to 0

, so the fringes crowd so tightly they blur into uniform illumination — the pattern effectively disappears.

Edge case: What is

Δ

and the brightness exactly at the central point

O

(

y = 0

Δ = 0

for all wavelengths, so it's the bright central maximum; in white light this is the only pure-white fringe since every colour agrees there.

Edge case: In white light, why is only the central fringe white while the sides show colours?

Away from centre, each colour has its own

β \propto λ

, so different colours peak at different

y

and separate out; only at

y = 0

do all colours coincide.

Edge case: What happens at extreme

y

where

y

is no longer

≪ D

The small-angle approximation fails, so

sin θ \neq = tan θ

; fringes become unequally spaced and gradually fainter as we leave the paraxial region.

Edge case: If a thin transparent sheet covers only slit

S_{1}

, what happens to the central fringe?

The sheet adds extra optical path to one arm, so the point of zero total path difference shifts sideways — the whole pattern (including the central bright fringe) moves off centre.