Young's double slit — fringe width derivation: Exercises

Figure — Young's double slit — fringe width derivation

The figure above re-anchors the picture: two slits $S_{1}, S_{2}$ a gap $d$ apart, a screen a distance $D$ away, and a point $P$ a height $y$ above centre. The pale-yellow stripes are the bright fringes, spaced $β$ apart.

Level 1 — Recognition

Recall Solution L1·Q1

WHAT: we want $β$ , spacing of consecutive bright fringes. WHY this formula: we know $λ, D, d$ directly, so use $β = \frac{λ D}{d}$ . Convert to SI first ( $1 nm = 1 0^{- 9} m$ , $1 mm = 1 0^{- 3} m$ ) so the answer is in metres: $β = \frac{( 500 \times 1 0 ^{- 9} ) ( 1.0 )}{0.25 \times 1 0 ^{- 3}} = \frac{5 \times 1 0 ^{- 7}}{2.5 \times 1 0 ^{- 4}} = 2 \times 1 0^{- 3} m = 2 mm$ Answer: $β = 2 mm$ .

Recall Solution L1·Q2

WHAT: classify $P$ . WHY: the bright/dark rule is stated purely in terms of $Δ$ .

Bright needs $Δ = nλ$ (a whole number of wavelengths).
Dark needs $Δ = (n + \frac{1}{2}) λ$ (a half-integer number). Here $Δ = 1.5 λ = (1 + \frac{1}{2}) λ$ , a half-integer multiple ⇒ dark fringe. Answer: dark.

Level 2 — Application

Recall Solution L2·Q1

WHY $y_{n}$ , not $β$ : we're told a specific fringe's position ( $n = 4$ ), so use $y_{n} = \frac{nλ D}{d}$ , then solve for $λ$ : $λ = \frac{y _{n} d}{n D} = \frac{( 5.6 \times 1 0 ^{- 3} ) ( 0.3 \times 1 0 ^{- 3} )}{4 \times 0.6} = \frac{1.68 \times 1 0 ^{- 6}}{2.4} = 7 \times 1 0^{- 7} m$ Answer: $λ = 700 nm$ (red light).

Recall Solution L2·Q2

WHAT: number of bright bands in a window of half-width $4.5 mm$ on each side of centre. Bright fringes sit at $y_{n} = n β$ . We need $∣ n β ∣ \leq 4.5 mm$ : $∣ n ∣ \leq \frac{4.5}{1.2} = 3.75 \Rightarrow n = - 3, - 2, - 1, 0, 1, 2, 3$ That's $7$ values. Answer: $7$ bright fringes.

Level 3 — Analysis

Recall Solution L3·Q1

WHY ratios: $β = \frac{λ D}{d}$ , so $β \propto \frac{D}{d}$ at fixed $λ$ . Track the multiplying factors, not the raw numbers: $\frac{β _{1}}{β _{0}} = \frac{D _{1} / d _{1}}{D _{0} / d _{0}} = \frac{( 3 D _{0} ) / ( d _{0} /2 )}{D _{0} / d _{0}} = 3 \times 2 = 6$ $β_{1} = 6 \times 1.5 mm = 9 mm$ Answer: $β_{1} = 9 mm$ — six times wider.

Recall Solution L3·Q2

WHAT: a point where $λ_{1}$ 's $n_{1}$ -th bright fringe lands on $λ_{2}$ 's $n_{2}$ -th bright fringe. WHY equate $y_{n}$ : overlap means the positions match: $\frac{n _{1} λ _{1} D}{d} = \frac{n _{2} λ _{2} D}{d} \Rightarrow n_{1} λ_{1} = n_{2} λ_{2} \Rightarrow \frac{n _{1}}{n _{2}} = \frac{λ _{2}}{λ _{1}} = \frac{600}{480} = \frac{5}{4}$ Smallest whole numbers: $n_{1} = 5, n_{2} = 4$ . The coincidence height: $y = \frac{n _{1} λ _{1} D}{d} = \frac{5 ( 480 \times 1 0 ^{- 9} ) ( 1.0 )}{0.5 \times 1 0 ^{- 3}} = \frac{2.4 \times 1 0 ^{- 6}}{5 \times 1 0 ^{- 4}} = 4.8 \times 1 0^{- 3} m$ Answer: $y = 4.8 mm$ (which is the 5th fringe of blue = 4th of orange).

Level 4 — Synthesis

Recall Solution L4·Q1

WHY $λ$ shrinks: frequency is fixed by the source; in a medium the wave slows by factor $n$ , so its wavelength shrinks: $λ^{'} = λ / n$ . (See Refractive index.) Since $β \propto λ$ and $D, d$ are unchanged: $β^{'} = \frac{λ ^{'} D}{d} = \frac{β _{air}}{n} = \frac{1.8}{1.5} = 1.2 mm$ Physical picture: a shorter ruler means the path-difference milestones $λ, 2 λ, \dots$ come up sooner, so bright bands crowd closer. Answer: $β^{'} = 1.2 mm$ .

Recall Solution L4·Q2

WHAT the sheet does: it makes the $S_{1}$ path optically longer by $(n - 1) t$ . The central fringe is where the total path difference is zero, so the geometric path must now favour $S_{1}$ by that same amount. Setting geometric path difference equal to the extra optical length: $\frac{y _{0} d}{D} = (n - 1) t$ WHY: $Δ_{geometry} = \frac{y _{0} d}{D}$ must cancel the added $(n - 1) t$ . $y_{0} = \frac{( n - 1 ) t D}{d} = \frac{( 0.5 ) ( 10 \times 1 0 ^{- 6} ) ( 1.2 )}{0.6 \times 1 0 ^{- 3}} = \frac{6 \times 1 0 ^{- 6}}{6 \times 1 0 ^{- 4}} = 1.0 \times 1 0^{- 2} m$ Answer: $y_{0} = 10 mm$ , shifted toward the side of the covered slit $S_{1}$ .

Level 5 — Mastery

Recall Solution L5·Q1

WHY ratios again: with $d, D$ fixed, $β \propto λ$ , so $\frac{λ _{X}}{λ _{R}} = \frac{β _{X}}{β _{R}} = \frac{1.5}{2.1} = \frac{5}{7}$ $λ_{X} = \frac{5}{7} \times 700 = 500 nm$ Answer: $λ_{X} = 500 nm$ (green).

Recall Solution L5·Q2

Step 1 — how many fringes each side. $9$ bright fringes symmetric about the centre means one central fringe ( $n = 0$ ) plus $4$ on each side ( $n = \pm 1, \dots, \pm 4$ ). So the outermost is $n = 4$ at $y_{4} = 6 mm$ . Step 2 — relate to $β$ . Since $y_{4} = 4 β$ : $β = \frac{y _{4}}{4} = \frac{6 mm}{4} = 1.5 mm$ Step 3 — solve for $d$ from $β = \frac{λ D}{d}$ : $d = \frac{λ D}{β} = \frac{( 546 \times 1 0 ^{- 9} ) ( 1.5 )}{1.5 \times 1 0 ^{- 3}} = \frac{8.19 \times 1 0 ^{- 7}}{1.5 \times 1 0 ^{- 3}} = 5.46 \times 1 0^{- 4} m$ Answer: $d \approx 0.546 mm$ .

Quick self-test

Fringe width formula?

β = \frac{λ D}{d}

Extra optical path added by a slab of thickness

t

, index

n

(n - 1) t

For two colours to coincide, what must be equal?

n_{1} λ_{1} = n_{2} λ_{2}

Outer bright order when

N

fringes fit symmetrically?

(N - 1) /2

Effect of submerging setup in liquid index

n

β

β \to β / n

Connections

Young's double slit — fringe width derivation — the parent derivation these problems drill.
Path difference and phase difference — L1·Q2 and L4·Q2 rest on this.
Refractive index — powers the immersion and slab problems (L4).
Small angle approximation — every $y = D θ$ step assumes it.
Interference of light · Coherence and coherent sources — the physics that makes fringes exist.