Wave Optics · Study notes

PHYInterference & Young's Double Slit

YDSE — Fringe Width & Fringe Positions

Young's double-slit experiment (YDSE) demonstrates the wave nature of light: two coherent slits separated by $d$ produce, on a screen a distance $D$ away, an alternating pattern of equally spaced bright and dark fringes. This article covers how to locate those fringes and how to compute the fringe width $\beta$ — the constant spacing between consecutive bright (or consecutive dark) fringes — together with the angular fringe width.

These are among the most reliably examined ISC numericals because they reduce to a single linear relation, $\beta = \dfrac{\lambda D}{d}$ , that can be rearranged to solve for any one of $\lambda$ , $d$ , $D$ , or a fringe position.

Fringe width (bright or dark)

\beta = \dfrac{\lambda D}{d}

\beta

= spacing between adjacent bright (or adjacent dark) fringes,

\lambda

= wavelength of light,

D

= slit-to-screen distance,

d

= separation between the two slits.

Position of the n-th bright fringe

y_n = \dfrac{n\lambda D}{d} = n\beta

y_n

= distance of the

n

-th bright fringe from the central maximum,

n = 0, 1, 2, \dots

(the central maximum is

n = 0

). All distances are measured on the same side of the centre.

Position of the n-th dark fringe

y_n = (2n-1)\dfrac{\lambda D}{2d} = (2n-1)\dfrac{\beta}{2}

y_n

= distance of the

n

-th dark fringe from the centre, with

n = 1, 2, 3, \dots

for the 1st, 2nd, 3rd,

\dots

dark fringe. Equivalent to path difference

(2n-1)\dfrac{\lambda}{2}

Separation between two bright fringes

\Delta y = y_p - y_q = (p-q)\dfrac{\lambda D}{d}

y_p, y_q

= positions of the

p

-th and

q

-th bright fringes on the same side of the centre; their gap equals

(p-q)

fringe widths, i.e.

(p-q)\beta

Angular fringe width

\theta = \dfrac{\beta}{D} = \dfrac{\lambda}{d}

\theta

= angular separation (in radians) between adjacent fringes as seen from the slits; it is independent of the screen distance

D

The pattern requires two coherent sources (constant phase relationship) that are monochromatic and of comparable intensity; coherence keeps the fringes stationary and visible, while equal intensities make the dark fringes truly dark (best contrast).
Fringe width $\beta$ is the same for bright and dark fringes and is constant across the pattern, so the fringes are equally spaced — this holds under the small-angle approximation valid near the centre of the screen.
Small-angle approximation: for $y \ll D$ the path difference is $\dfrac{yd}{D}$ , so $\sin\theta \approx \tan\theta \approx \theta = \dfrac{y}{D}$ (with $\theta$ in radians). Every standard fringe formula rests on this.
Bright fringes (maxima) occur where the path difference $= n\lambda$ ; dark fringes (minima) where it $= (2n-1)\dfrac{\lambda}{2}$ . The central maximum ( $n = 0$ ) is bright and lies at $y = 0$ .
Dependence of fringe width: $\beta \propto \lambda$ , $\beta \propto D$ , and $\beta \propto \dfrac{1}{d}$ . So increasing the slit separation $d$ shrinks the fringes, while increasing $D$ or using longer-wavelength (redder) light widens them.
Angular fringe width $\theta = \dfrac{\lambda}{d}$ does not depend on $D$ : moving the screen changes the linear fringe width $\beta$ but not the angular spacing.
To find the gap between the $p$ -th and $q$ -th bright fringes on the same side, take $(p-q)\beta$ ; if the two fringes lie on opposite sides of the centre, add their distances from the centre instead of subtracting.
Always convert every length to one consistent unit (SI metres is safest) before substituting: $1\,\text{nm} = 10^{-9}\,\text{m}$ , $1\,\text{mm} = 10^{-3}\,\text{m}$ .

Where the marks go

Mixing units: leaving $\lambda$ in nm while $d$ is in mm and $D$ in m throws the answer off by large powers of ten. Convert all lengths to metres first, then convert the final answer back to mm if required.
Confusing position with spacing: the 5th bright fringe is at $y_5 = 5\beta$ , but the distance between the 1st and 5th bright fringes is $(5-1)\beta = 4\beta$ , not $5\beta$ .
Mismatching the formula: using the dark-fringe expression $(2n-1)\dfrac{\beta}{2}$ for a bright fringe (or vice versa). Match the formula to whether the question asks for a maximum or a minimum, and remember the central bright fringe is $n = 0$ while the first dark fringe is $n = 1$ .
Treating angular fringe width as dependent on $D$ : it is $\theta = \dfrac{\lambda}{d}$ only — do not multiply or divide by the screen distance, and keep $\theta$ in radians unless degrees are explicitly asked for.

How the board asks it

Numericalfringe-width relation $\beta = \dfrac{\lambda D}{d}$
In a Young's double-slit experiment, two slits separated by $d = 0.8\,\text{mm}$ are illuminated by light of wavelength $\lambda = 589\,\text{nm}$ , and the fringes are observed on a screen $D = 1.2\,\text{m}$ away. Calculate the fringe width $\beta$ .
Numericalposition of the $n$ -th bright fringe and fringe separation
In a YDSE with $\lambda = 6000\,\text{\AA}$ , $d = 1\,\text{mm}$ and $D = 1\,\text{m}$ , calculate the distance of the $4$ th bright fringe from the centre and the separation between the $2$ nd and $5$ th bright fringes.
Numerical $\beta = \dfrac{\lambda D}{d}$ rearranged for $\lambda$
In a double-slit experiment the distance between the slits is $0.5\,\text{mm}$ and the screen is $1\,\text{m}$ away. If $10$ fringes occupy a width of $9\,\text{mm}$ , calculate the wavelength of the light used.
Derive / provesmall-angle approximation and path difference $\dfrac{yd}{D}$
With the help of a suitable diagram, derive an expression for the fringe width $\beta$ of the fringes obtained in Young's double-slit experiment.
Give reasonsdependence $\beta \propto \lambda$ , $\beta \propto D$ , $\beta \propto \dfrac{1}{d}$
In a YDSE, give reasons for how the fringe width changes when (i) the slit separation $d$ is increased, and (ii) the whole apparatus is immersed in water of refractive index $\mu = 1.33$ .
Define / stateangular fringe width $\theta = \dfrac{\lambda}{d}$
Define the fringe width in Young's double-slit experiment. State, with reason, whether the angular fringe width $\theta = \dfrac{\lambda}{d}$ changes when the screen distance $D$ is increased.

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Written for Sublevo. Question text quoted anywhere in these notes is the Council’s and carries its year and paper; the board’s own diagrams are not reproduced.