Path Difference, Slab Shift & Immersion

PHYInterference & Young's Double Slit

In Young's double-slit experiment the position and nature (bright or dark) of any point on the screen is decided entirely by the optical path difference $\Delta x$ between the two interfering beams. This article covers three classic ways that path difference is manipulated: a point displaced from the central maximum ( $\Delta x = \dfrac{yd}{D}$ ), immersing the whole set-up in a medium (which shrinks the wavelength and hence the fringe width), and introducing a thin transparent slab in one beam (which adds an extra optical path $(\mu-1)t$ and shifts the entire pattern sideways).

ISC examiners favour these because a single small-angle relation plus the bright/dark condition lets them test path difference, fringe width and pattern shift in one compact numerical.

Path difference at point P

\Delta x = \dfrac{y\,d}{D} \approx d\sin\theta

y

is the distance of P from the central maximum,

d

the slit separation,

D

the slit-to-screen distance, and

\theta

the angular position; valid for

y \ll D

(small angles), where

\sin\theta \approx \tan\theta = \dfrac{y}{D}

Bright / dark condition

\Delta x = n\lambda \;(\text{bright}),\qquad \Delta x = \left(n-\tfrac{1}{2}\right)\lambda \;(\text{dark})

\lambda

is the wavelength; a point is bright when

\Delta x

is an integer multiple of

\lambda

(

n=0,1,2,\dots

) and dark when it is an odd multiple of

\dfrac{\lambda}{2}

(

n=1,2,3,\dots

in the form above).

Fringe width

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

\beta

is the spacing between adjacent bright (or dark) fringes; the position of the

n

th bright fringe follows from this through

y_n = n\beta

Fringe width in a medium (immersion)

\beta_{m} = \dfrac{\lambda_{m} D}{d} = \dfrac{\beta}{\mu}

\mu

is the refractive index of the medium filling the apparatus and

\lambda_{m} = \dfrac{\lambda}{\mu}

is the wavelength inside it; the fringes get closer because the wavelength shrinks.

Extra optical path from a thin slab

\Delta x_{slab} = (\mu - 1)\,t

\mu

is the refractive index of the slab and

t

its thickness; this is the extra optical path the beam through the slab gains compared with travelling the same geometric distance in air.

Lateral shift of the pattern

S = \dfrac{(\mu-1)\,t\,D}{d} = N\beta, \quad N = \dfrac{(\mu-1)t}{\lambda}

S

is the sideways displacement of the whole fringe system (towards the slab side), and

N

is the number of fringes shifted; note

N

is independent of

D

and

d

The small-angle relation $\Delta x = \dfrac{yd}{D}$ replaces the exact $d\sin\theta$ only when $y \ll D$ ; in standard ISC set-ups ( $D \sim 1\,\text{m}$ , $y \sim \text{mm}$ ) this is excellent, so $\sin\theta \approx \tan\theta = \dfrac{y}{D}$ .
To classify a point, compute $\dfrac{\Delta x}{\lambda}$ : an integer means bright, a half-integer (like $2.5$ ) means dark. This single ratio is faster and safer than memorising separate $y$ -formulae.
On immersion only the wavelength changes ( $\lambda_m = \lambda/\mu$ ); the geometry $D$ and $d$ are unchanged, so $\beta$ scales as $1/\mu$ and the pattern simply gets finer, never wider.
Frequency does not change on entering a medium; only speed and wavelength do, which is why we use $\lambda_m = \lambda/\mu$ rather than altering the value of $\lambda$ in air.
The slab's effect uses $(\mu-1)t$ , not $\mu t$ : the beam already would have travelled a geometric path $t$ in air, so only the extra optical path $(\mu-1)t$ counts.
The lateral shift $S=\dfrac{(\mu-1)tD}{d}$ moves the entire pattern towards the side carrying the slab, but it does not change the fringe width $\beta$ — fringe spacing depends only on $\lambda$ , $D$ and $d$ .
The number of fringes shifted, $N=\dfrac{(\mu-1)t}{\lambda}$ , is dimensionless and independent of $D$ and $d$ ; equate $S = N\beta$ as a quick consistency check.
With white light the central fringe (zero path difference) is white and surrounding fringes are coloured; a slab makes this central white fringe move to where the extra path is exactly cancelled — monochromatic-only formulae must not be quoted for white-light parts.

Where the marks go

Unit slips: $d$ , $D$ , $y$ and $t$ often arrive in mm, m and $\mu\text{m}$ while $\lambda$ is in nm. Convert everything to metres first — e.g. $t=10\,\mu\text{m}=10\times10^{-6}\,\text{m}$ , $\lambda=500\,\text{nm}=500\times10^{-9}\,\text{m}$ — before substituting.
Using $\mu t$ instead of $(\mu-1)t$ for the slab (the single most common slab error). Always subtract the air path that the beam would have covered anyway, so that an air slab ( $\mu=1$ ) correctly gives zero extra path.
Treating immersion as if it widens the fringes: the correct result divides by $\mu$ ( $\beta_m=\beta/\mu$ ), so fringes get closer; multiplying by $\mu$ instead pushes the answer in the wrong direction and overestimates the spacing.
Mis-reading $\Delta x = 2.5\lambda$ as bright: a half-integer multiple of $\lambda$ is dark. Equivalently $2.5\lambda=\left(3-\tfrac12\right)\lambda$ , an odd multiple of $\dfrac{\lambda}{2}$ , so it is the $n=3$ dark fringe — not bright.

How the board asks it

Numerical $\Delta x = \dfrac{yd}{D}$ and the bright/dark condition
In a double-slit experiment $d = 0.5\,\text{mm}$ , $D = 1.2\,\text{m}$ and $\lambda = 500\,\text{nm}$ . Calculate the path difference at a point $P$ which is $4.5\,\text{mm}$ from the central maximum, and state whether $P$ is bright or dark.
Numericalextra path $(\mu-1)t$ , lateral shift $S=\dfrac{(\mu-1)tD}{d}$ and $N=\dfrac{(\mu-1)t}{\lambda}$
A thin glass plate of refractive index $\mu = 1.5$ and thickness $t = 6\,\mu\text{m}$ is introduced in the path of one of the interfering beams in a Young's double-slit experiment. If $D = 1\,\text{m}$ , $d = 1\,\text{mm}$ and $\lambda = 600\,\text{nm}$ , calculate the lateral shift of the fringe pattern and the number of fringes by which it shifts.
Numericalfringe width in a medium, $\beta_m = \dfrac{\beta}{\mu}$
The fringe width in a Young's double-slit experiment is $0.4\,\text{mm}$ when the apparatus is in air. Calculate the new fringe width if the entire arrangement is immersed in water of refractive index $\mu = \dfrac{4}{3}$ .
Derive / provelateral shift from the extra optical path $(\mu-1)t$
Derive an expression for the lateral shift of the fringe pattern produced when a thin transparent slab of thickness $t$ and refractive index $\mu$ is placed in front of one of the slits in a Young's double-slit experiment.
Give reasonsimmersion shrinks $\lambda$ but not $D$ or $d$ ; a slab shifts the pattern without changing $\beta$
Give reasons: (i) when a Young's double-slit apparatus is immersed in a liquid the fringes become narrower; (ii) introducing a thin transparent slab in front of one slit shifts the whole pattern but leaves the fringe width $\beta$ unchanged.
Applicationwhite-light central fringe and slab cancellation
In a Young's double-slit experiment performed with white light, state what is observed at the centre of the screen, and explain how this central fringe shifts when a thin mica slab is placed in front of one slit.

Practise this topic

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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.