In a two-slit interference set-up a source $A$ sends light to two slits $S_1$ and $S_2$ whose midpoint is $C$, with $S_1C = S_2C = d$ (slit separation $2d$). The screen is at perpendicular distance $CO = D$ from the slit plane, with $O$ the point on the screen opposite $C$; on the source side $AC = D$, and $A$ lies on the axis so that $AS_1 = AS_2$. It is given that $d \ll D$. A thin transparent slab of refractive index $\mu = 1.5$ and thickness $L = d/4$ is inserted in the path $A \to S_2$ only. Given that without the slab the principal maximum is at $O$, find the distance from $O$ at which the principal maximum now appears and the distances from $O$ of the first minima on either side of it. Express the answers in terms of $D$, $d$ and the wavelength $\lambda$.

In a two-slit interference set-up a source A sends light to two slits S₁ and S₂ whose midpoint is C, with S₁C

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Concept understanding — Double Slit Interference

Double Slit Interference: From Ripples to Light

Imagine dropping two stones into a still pond at the same time, a short distance apart. Watch the ripples spread. Where a crest from one stone meets a crest from the other, the water rises higher. Where a crest meets a trough, the water flattens out. That is interference — waves adding or cancelling.

Now replace the water with light. Replace the stones with two narrow slits cut into a barrier. Shine a single colour of light (say, red laser light) onto the slits. On a screen behind the barrier, you do not see two bright spots. Instead, you see a pattern of alternating bright and dark bands — like a striped zebra crossing made of light.

That pattern is double slit interference. It is the single most convincing proof that light behaves as a wave.

The Core Idea

Light from a single source passes through two narrow slits. Each slit acts as a new source of waves. These two sets of waves spread out and overlap. At any point on the screen, the light you see is the sum of the waves from slit 1 and slit 2.

Whether they add (bright) or cancel (dark) depends on one thing: the path difference — how much farther one wave has travelled compared to the other.

Note

For constructive interference (bright band): path difference = $n\lambda$ (whole number of wavelengths)

For destructive interference (dark band): path difference = $(n + \frac12)\lambda$ (half-integer number of wavelengths)

Here $\lambda$ is the wavelength of the light, and $n = 0, 1, 2, \dots$

The Geometry

Let the slits be separated by distance $d$ . The screen is far away at distance $D$ ( $D \gg d$ ). For a point on the screen at angle $\theta$ from the centre:

The path difference $\Delta x = d \sin\theta$
Bright bands occur when $d \sin\theta = n\lambda$
Dark bands occur when $d \sin\theta = (n + \frac12)\lambda$

The position $y$ of the $n$ -th bright band on the screen (measured from the centre) is:

y_n = \frac{n\lambda D}{d}

The spacing between consecutive bright bands (fringe width) is:

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

\text{Fringe width } \beta = \frac{\lambda D}{d}

What This Tells You

Larger $\lambda$ → wider fringes (red light spreads more than blue)
Larger $D$ → wider fringes (screen further away spreads the pattern)
Smaller $d$ → wider fringes (slits closer together spread the pattern more)

If you cover one slit, the pattern vanishes — you get a single blurry blob. The stripes only appear when both slits are open, proving that the light from the two slits is interfering.

Why It Matters

Double slit interference is not a classroom toy. It is the foundation of:

Young's experiment (1801) — which settled the debate: light is a wave
Diffraction gratings — used in spectrometers to identify elements by their light …

Why this formula?

Double Slit Interference: Why the Formula Holds

Let's build the understanding from first principles — not just memorise the formula, but see why it must be true.

1. The Core Idea: Path Difference Creates Phase Difference

Imagine two narrow slits $S_1$ and $S_2$ , separated by distance $d$ , illuminated by a single coherent source. Light from each slit travels to a point $P$ on a screen at distance $D$ (where $D \gg d$ ).

The two waves start in phase at the slits (same source).
They travel different distances to reach $P$ .
This path difference $\Delta x$ causes a phase difference $\Delta \phi$ .

Key relation:

\Delta \phi = \frac{2\pi}{\lambda} \cdot \Delta x

Why? Because one full wavelength $\lambda$ corresponds to a phase change of $2\pi$ radians.

2. Finding the Path Difference

From the geometry (see diagram in any textbook):

For a point $P$ at angle $\theta$ from the central axis, the extra distance travelled by the wave from the farther slit is approximately:

\Delta x = d \sin\theta

Why approximate? Because we assume $D \gg d$ , so the two paths are nearly parallel. This is the Fraunhofer (far-field) approximation — valid for most exam setups.

3. Condition for Constructive Interference (Bright Fringes)

Waves interfere constructively when they arrive in phase:

\Delta \phi = 2\pi m \quad (m = 0, \pm1, \pm2, \dots)

Using $\Delta \phi = \frac{2\pi}{\lambda} \cdot d\sin\theta$ , we get:

\frac{2\pi}{\lambda} \cdot d\sin\theta = 2\pi m

Cancel $2\pi$ to obtain the bright fringe condition:

\boxed{d \sin\theta = m\lambda}

$m$ is called the order of the fringe.
$m=0$ gives the central bright fringe (straight ahead).

4. Condition for Destructive Interference (Dark Fringes)

Waves interfere destructively when they arrive out of phase by $\pi$ (half a cycle):

\Delta \phi = (2m+1)\pi

Substitute again:

\frac{2\pi}{\lambda} \cdot d\sin\theta = (2m+1)\pi

Cancel $\pi$ to get the dark fringe condition:

\boxed{d \sin\theta = \left(m + \frac12\right)\lambda}

5. From Angle to Position on Screen

For small angles (typical in exam problems), $\sin\theta \approx \tan\theta = \frac{y}{D}$ , where $y$ is the distance from the central maximum on the screen.

Bright fringe position:

y_m = \frac{m\lambda D}{d}

Dark fringe position:

$y_m = \frac{(m + \frac12)\lambda D}{d}$ …

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Inserting the slab in the $S_2$ path lengthens that arm's optical path by $(\mu-1)L=d/8$ , which slides the entire fringe pattern by $D/16$ toward the $S_2$ side. The principal maximum therefore sits $D/16$ from $O$ , and the first minima lie half a fringe width ( $\lambda D/4d$ ) on each side of it.

Extra optical path from the slab

A slab of index $\mu$ and thickness $L$ replaces a length $L$ of air, adding optical path

\delta=(\mu-1)L=(1.5-1)\cdot\frac{d}{4}=\frac12\cdot\frac{d}{4}=\frac{d}{8}.

This extra path is in the $S_2$ arm, so the wave through $S_2$ now lags.

Total path difference at a point $P$ (height $y$ above $O$ )

Since $AS_1=AS_2$ , the source contributes no path difference. With slit separation $2d$ and screen distance $D$ (and $d\ll D$ ),

\Delta(y)=\underbrace{\frac{2d\,y}{D}}_{\text{geometry}}+\underbrace{\frac{d}{8}}_{\text{slab}},

taking $y$ positive toward $S_1$ (so the slab term and geometry term have opposite effect on the $S_2$ -side).

Principal (zero-order) maximum: $\Delta=0$

\frac{2d\,y}{D}+\frac{d}{8}=0\;\Rightarrow\; y=-\frac{D}{16}.

The negative sign means the maximum shifts a distance $\dfrac{D}{16}$ toward the $S_2$ (slab) side of $O$ . Equivalently, using the fringe-shift formula $\Delta y=\dfrac{(\mu-1)L\,D}{2d}=\dfrac{(d/8)D}{2d}=\dfrac{D}{16}.$

First minima: $\Delta=\pm\dfrac{\lambda}{2}$ …

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Same / Similar Concept — real previous-year questions on the same or a closely similar concept, not this exact question.

Showing the 12 most recent of 13 on this concept.

GUJCET 2026Set x1 markMCQ
$\text{\AA}$
›Reveal solutionSolution
[!TLDR] Substituting the given numbers into $\lambda=\beta d/D$ yields $1.5\times10^{-6}$ m $=15000\,\text{\AA}$ , matching option (D).

Concept

In Young's double-slit experiment the fringe width is $\beta = \dfrac{\lambda D}{d}$ , where $d$ is the slit separation and $D$ the slit-to-screen distance. Rearranged, $\lambda = \dfrac{\beta d}{D}$ .

Solution

Given $d = 0.2\,\text{mm} = 0.2\times10^{-3}$ m, $D = 2.0$ m, $\beta = 1.5\,\text{cm} = 1.5\times10^{-2}$ m. …
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GSEB Higher Secondary Certificate (HSC) Examination 2026Set ANNUAL1 markMCQ
$Q. If path difference between two waves is 3 lambda then phase difference related with them is ___.$
(a) 4 pi rad
(b) 3 pi rad
(c) 2 pi rad
(d) 6 pi rad
›Reveal solutionSolution

Path difference and phase difference are related by delta(phi) = (2 pi/lambda) x (path difference), because one full wavelength of path corresponds to a full 2 pi cycle of phase.

…

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GUJCET 2025Set 031 markMCQ
$I_{0}$
›Reveal solutionSolution
Superposition of two equal-intensity coherent waves gives $I=4I_0\cos^2(\phi/2)$ .

Concept — interference. Resultant intensity of two waves of intensities $I_1,I_2$ with phase difference $\phi$ is $I=I_1+I_2+2\sqrt{I_1I_2}\cos\phi$ .

Steps.

With $I_1=I_2=I_0$ : $I=2I_0+2I_0\cos\phi=2I_0(1+\cos\phi)$ . …
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GUJCET 2025Set 031 markMCQ
$Q. In Young's double slit experiment, the slits are separated by 0.54 mm and the screen is placed 1.8 m away. The distance between central bright fringe and sixth bright fringe is measured to be 1.2 cm. Determine the wavelength of light used in the experiment. (A) 5000 Å (B) 600 nm (C) 8000 nm (D) 800 nm$
›Reveal solutionSolution
[!TLDR]

From $\lambda = x_n d/(nD)$ the wavelength is $600\,\text{nm}$ .

Concept

In Young's double-slit experiment the distance of the $n$ -th bright fringe from the centre is $x_n = \dfrac{n\lambda D}{d}$ , so $\lambda = \dfrac{x_n\,d}{n\,D}$ .

Solution

Given $d = 0.54\,\text{mm} = 0.54\times10^{-3}\,\text{m}$ , $D = 1.8\,\text{m}$ , $x_6 = 1.2\,\text{cm} = 1.2\times10^{-2}\,\text{m}$ , $n = 6$ : …
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GUJCET 2024Set 131 markMCQ
$Q. In a Young's double-slit experiment, the slits are separated by 0.28 mm and the screen is placed 1.4 m away. The distance between the central bright fringe and the fourth bright fringe is measured to be 1.2 cm. Then the wavelength of light used in the experiment is ________. (A) 500 nm (B) 660 nm (C) 600 nm (D) 550 nm$
›Reveal solutionSolution

Fourth bright fringe: $y_4 = \frac{4\lambda D}{d}$ , solve for $\lambda$ .

Steps. $d=0.28\times10^{-3}$ m, $D=1.4$ m, $y_4 = 1.2\times10^{-2}$ m, $n=4$ . …
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GUJCET 2023Set 091 markMCQ
$7000\,\text{\AA}$
›Reveal solutionSolution
[!TLDR]

Using $\beta = \lambda D/d$ gives a fringe width of $105\,\mu m$ .

Concept

In Young's double-slit experiment the spacing between adjacent bright (or dark) fringes is $\beta = \dfrac{\lambda D}{d}$ , where $D$ is the slit-to-screen distance and $d$ the slit separation.

Solution

Convert units: $\lambda = 7000\,\text{\AA} = 7\times10^{-7}\,\text{m}$ , $D = 1.5\,\text{m}$ , $d = 10\,\text{mm} = 1\times10^{-2}\,\text{m}$ . …
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GSEB Higher Secondary Certificate (HSC) Examination 2023Set ANNUAL1 markMCQ
$Q. In Young's double experiment the distance between two slits is 0.2 mm and the distance between slit and screen is 1.5 m. The wavelength of light used is 600 nm. The distance between any two consecutive bright fringes is ___ mm.$
(a) 0.8
(b) 4.5
(c) 0.5
(d) 2.0
›Reveal solutionSolution

The spacing of consecutive bright fringes is beta = lambda D/d = 4.5 mm.

Fringe width in Young's double-slit experiment:

beta = lambda D / d.

Substitute lambda = 600 nm = 600 x 10^-9 m, D = 1.5 m, d = 0.2 mm = 0.2 x 10^-3 m: …

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GUJCET 2022Set 171 markMCQ
$Q. Two slits are made 3 millimetre (3 mm) apart and the screen is placed 2 m away. What is the fringe separation when blue-green light of wavelength 600 nm is used? (A) 0.4 mm (B) 0.6 mm (C) 0.5 mm (D) 0.7 mm$
›Reveal solutionSolution
$\beta = \dfrac{\lambda D}{d}$ .

Steps.

$\lambda = 600\text{ nm} = 6\times10^{-7}$ m, $D = 2$ m, $d = 3\text{ mm} = 3\times10^{-3}$ m. …
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GUJCET 2021Set 151 markMCQ
$Q. The wavelength of light 500 nm is used in a Young's double-slit experiment. The distance between the slits and screen is 100 cm and the slits are separated by 1 mm. Then find distance between fifth (5th) and third (3rd) bright fringes. (A) 1 mm (B) 3 mm (C) 2 mm (D) 4 mm$
›Reveal solutionSolution

Consecutive bright fringes are beta = lambda D/d apart; the 5th and 3rd differ by 2 beta.

Concept. $\beta=\dfrac{\lambda D}{d}$ , and $x_5-x_3=(5-3)\beta=2\beta$ . …
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GUJCET 2020Set 071 markMCQ
$Q. The distance between two slits is 3 mm & screen is placed at 2 m distance. When blue-green light of wavelength 500 nm is used then distance between two fringes will be? (A) 0.5 mm (B) 0.43 mm (C) 0.33 mm (D) 0.4 mm$
›Reveal solutionSolution

Fringe spacing in YDSE is $\beta=\dfrac{\lambda D}{d}$ .

Concept: In Young's double-slit experiment the distance between adjacent bright (or dark) fringes is $\beta=\dfrac{\lambda D}{d}$ . …
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GUJCET 2019Set 131 markMCQ
$5000\,\text{Å}$
›Reveal solutionSolution

Overlapping bright fringes have equal positions, so $n_1\lambda_1 = n_2\lambda_2$ .

Concept first: The position of the $n$ -th bright fringe is $y_n = n\lambda D/d$ . If the 4th bright fringe of $5000\,\text{Å}$ coincides with the 5th bright fringe of the unknown wavelength: …
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GUJCET 2015Set C1 markMCQ
$\lambda$
›Reveal solutionSolution
[!TLDR] Setting the central-maximum width equal to the slit width gives $D = \dfrac{d^2}{2\lambda}$ , option (B).

Concept …
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Extra optical path from the slab

Total path difference at a point PPP (height yyy above OOO)

Principal (zero-order) maximum: Δ=0\Delta=0Δ=0

First minima: Δ=±λ2\Delta=\pm\dfrac{\lambda}{2}Δ=±2λ​ …

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Total path difference at a point $P$ (height $y$ above $O$ )

Principal (zero-order) maximum: $\Delta=0$

First minima: $\Delta=\pm\dfrac{\lambda}{2}$ …

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