In a two-slit interference arrangement the slits $S_1$ and $S_2$ are separated by a distance $2D$; their midpoint is $C$ so that $S_1C = CS_2 = D$. A screen is placed parallel to the slit plane at a perpendicular distance $CO = D$ from the slits, where $O$ is the point on the screen directly opposite $C$. Hence the slit-to-screen distance ($D$) equals half the slit separation ($2D$). Monochromatic light of wavelength $\lambda$ is used. Obtain the value of $D$ in terms of $\lambda$ such that the first minimum on the screen falls at a point $P$ for which the distance $OP$ from the centre $O$ is equal to $D$.

In a two-slit interference arrangement the slits S₁ and S₂ are separated by a distance 2D; their midpoint is C

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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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Because the screen is close to the slits (distance $D$ against a slit separation $2D$ ), the small-angle approximation cannot be used — the exact path lengths must be computed. Placing the point $P$ at $OP=D$ and demanding the exact path difference equal $\lambda/2$ (first minimum) gives $D=\dfrac{(\sqrt5+1)\lambda}{8}\approx0.404\lambda.$

Set up coordinates

Take $C$ as the origin, the slit plane along the $y$ -axis and the screen along $x=D$ :

S_1=(0,\;D),\qquad S_2=(0,\;-D),\qquad O=(D,\;0).

The required point $P$ is on the screen with $OP=D$ , so $P=(D,\;D)$ .

Exact path lengths

S_1P=\sqrt{(D-0)^2+(D-D)^2}=\sqrt{D^2}=D,

S_2P=\sqrt{(D-0)^2+(D+D)^2}=\sqrt{D^2+4D^2}=\sqrt5\,D.

Path difference and the minimum condition

$\Delta = S_2P-S_1P=\sqrt5\,D-D=(\sqrt5-1)\,D.$ …

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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 18 on this concept.

KEAM 2026Set eng-2026-04174 marksMCQ
$Q. The width of a fringe is 0.5 mm in Young's double slit experiment for a light of wavelength 500 nm. If the wave length of light alone is changed to 600 nm, the width of the fringe becomes (A) 0.4 mm (B) 0.3 mm (C) 0.2 mm (D) 0.6 mm (E) 0.55 mm$
›Reveal solutionSolution

In Young's experiment $\beta = \dfrac{\lambda D}{d}$ , so with only $\lambda$ changed the fringe width scales directly with wavelength.

Given $\beta_1 = 0.5$ mm at $\lambda_1 = 500$ nm. With $D$ and $d$ unchanged: …
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KEAM 2026Set eng-2026-04184 marksMCQ
$1m$
›Reveal solutionSolution

Fringe width $\beta=\lambda D/d$ , so $\Delta\beta=\Delta\lambda\,D/d$ ; solving gives $d=0.4\,mm$ .

Fringe width $\beta=\dfrac{\lambda D}{d}$ , so a change in wavelength changes it by

$\Delta\beta=\frac{\Delta\lambda\,D}{d}\ \Rightarrow\ d=\frac{\Delta\lambda\,D}{\Delta\beta}$ …
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KEAM 2026Set eng-2026-04204 marksMCQ
$Q. Sustained interference is observed with (A) two independent sources (B) two coherent sources (C) a single coherent source (D) a single independent source (E) a single monochromatic source$
›Reveal solutionSolution

A steady interference pattern needs two coherent sources maintaining a constant phase difference.

Reasoning. Independent sources have randomly varying phases, washing out the pattern. Sustained interference is observed only when the two interfering waves come from coherent sources with …
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KEAM 2026Set eng-2026-04214 marksMCQ
$\beta_R$
›Reveal solutionSolution
[!TLDR]

Because fringe width $\beta \propto \lambda$ and $\lambda_R > \lambda_G > \lambda_B$ , the correct relation is $\beta_R > \beta_G$ , option (B).

Concept

This is from the NCERT/CBSE Wave Optics chapter. In Young's double-slit experiment the spacing between adjacent bright (or dark) fringes is
$\beta = \frac{\lambda D}{d},$
where $D$ is the slit-to-screen distance and $d$ the slit separation. For a fixed setup ( $D$ and $d$ constant), $\beta$ is directly proportional to the wavelength $\lambda$ .

Solution

The wavelengths satisfy …
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KEAM 2026Set eng-2026-04224 marksMCQ
$Q. If two waves of same wavelength with their intensities in the ratio 25 : 9 produce interference, then the ratio of the maximum to minimum intensity is (A) 9 : 2 (B) 9 : 1 (C) 5 : 3 (D) 16 : 3 (E) 16 : 1$
›Reveal solutionSolution

Intensity $\propto$ amplitude $^2$ . With $I_1:I_2=25:9$ the amplitudes are $5:3$ , giving $I_{max}:I_{min}=(a_1+a_2)^2:(a_1-a_2)^2=16:1$ .

Since $I\propto a^2$ , the amplitude ratio is
$\frac{a_1}{a_2}=\sqrt{\frac{25}{9}}=\frac{5}{3}.$
In interference, …
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KEAM 2026Set pha-2026-0419F4 marksMCQ
$Q. In Young's double slit experiment, monochromatic light of 500 nm falls on the slits separated by a distance of 2 mm. If the screen is 2 m away from the source, then the fringe width is (A) 5 mm (B) 2.0 mm (C) 0.5 mm (D) 2.5 mm (E) 1.5 mm$
›Reveal solutionSolution

Fringe width $\beta = \lambda D/d = (500\ \mathrm{nm})(2\ \mathrm{m})/(2\ \mathrm{mm}) = 0.5\ \mathrm{mm}$ .

The fringe width in Young's double-slit experiment is
$\beta = \frac{\lambda D}{d},$
with $\lambda = 500\times10^{-9}$ m, $D = 2$ m, $d = 2\times10^{-3}$ m: …
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KEAM 2025Set eng-2025-04234 marksMCQ
$\beta$
›Reveal solutionSolution
To double the fringe width the wavelength must be doubled to 2*lambda.

Concept and Intuition

Fringe width beta = lambda*D/d is directly proportional to wavelength when D and d are fixed. So doubling beta requires doubling lambda.

Step-by-Step Solution

beta = lambda*D/d, so beta proportional to lambda.

To make beta' = 2beta with D, d fixed, need lambda' = 2lambda. …
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KEAM 2025Set eng-2025-04254 marksMCQ
$\lambda$
›Reveal solutionSolution

In YDSE the angular fringe width is $\omega=\lambda/d$ , independent of screen distance. Halving the slit separation $d$ doubles $\omega$ .

The angular separation of fringes is

$\omega=\dfrac{\lambda}{d}.$ …
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KEAM 2025Set eng-2025-04284 marksMCQ
$A^{2}$
›Reveal solutionSolution

Two equal-amplitude waves in opposite phase cancel exactly: resultant amplitude $=0$ .

The resultant amplitude of two interfering waves is

$A_R=\sqrt{A_1^2+A_2^2+2A_1A_2\cos\phi}.$ …
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KEAM 2025Set eng-2025-04294 marksMCQ
$nm$
›Reveal solutionSolution

Fringe width $\beta=\dfrac{\lambda D}{d}$ is directly proportional to wavelength (with $D$ and $d$ fixed), so $\beta$ becomes $3\times$ when $\lambda$ is tripled. …

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KEAM 2025Set pha-2025-0424F4 marksMCQ
$Q.In Young's double slit experiment performed in air medium, the fringe width observed is 1.4 mm. If the entire arrangement is kept in a liquid medium of refractive index 1.4, then the fringe width (in mm) will be (A) 1.4 (B) 1.0 (C) 0.7 (D) 2.8 (E) 0.5$
›Reveal solutionSolution

In a medium of refractive index $n$ the wavelength (and hence fringe width) reduces by $n$ : $\beta'=\beta/n=1.4/1.4=1.0$ mm.

Fringe width is

$\beta=\frac{\lambda D}{d}.$ …
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KEAM 2024Set eng-2024-06054 marksMCQ
$Q. In an Young double slit experiment without varying the distance of the screen and the slit separation if the wavelength of monochromatic source is changed one by one in the ratio 2 : 3 : 4 then the corresponding fringe widths measured will be in the ratio (A) 4 : 3 : 2 (B) 1 : 2 : 3 (C) 2 : 3 : 4 (D) 6 : 4 : 3 (E) 3 : 4 : 6$
›Reveal solutionSolution

$\beta \propto \lambda$ ; with $D,d$ constant the fringe widths are in the same ratio $2:3:4$ .

In Young's double-slit experiment the fringe width is

$\beta = \frac{\lambda D}{d}.$ …
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Concept understanding — Double Slit Interference

Double Slit Interference: From Ripples to Light

The Core Idea

The Geometry

What This Tells You

Why It Matters

Why this formula?

Double Slit Interference: Why the Formula Holds

1. The Core Idea: Path Difference Creates Phase Difference

2. Finding the Path Difference

3. Condition for Constructive Interference (Bright Fringes)

4. Condition for Destructive Interference (Dark Fringes)

5. From Angle to Position on Screen

Set up coordinates

Exact path lengths

Path difference and the minimum condition

Method: Locating Interference Minima/Maxima Without the Small-Angle Approximation

Steps

Concept

Solution

Concept and Intuition

Step-by-Step Solution

Set up coordinates

Exact path lengths

Path difference and the minimum condition

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