What is the effect on the interference fringes in Young's double slit experiment due to each of the following

Question

Q. What is the effect on the interference fringes in Young's double slit experiment due to each of the following operations ? Justify your answers.

(a) The screen is moved away from the plane of the slits.

(b) The separation between slits is increased.

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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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The interference pattern in Young’s double slit experiment is governed by the fringe width $\beta = \frac{\lambda D}{d}$ . Moving the screen away ( $D \uparrow$ ) widens fringes; increasing slit separation ( $d \uparrow$ ) narrows fringes; moving the source slit closer does not change fringe width but reduces fringe contrast (visibility) because the two slits are no longer equally illuminated.

The core idea

Young’s double slit experiment creates an interference pattern because light from a single source slit reaches the two slits $S_1$ and $S_2$ in phase. These two slits then act as coherent sources. On a screen at distance $D$ , the bright and dark fringes are equally spaced. The fringe width (distance between consecutive bright or dark fringes) is given by:

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

where $\lambda$ is the wavelength of light, $D$ is the distance from the slits to the screen, and $d$ is the separation between the two slits.

This formula is the key to answering parts (a) and (b). Part (c) is different — it affects the quality of the pattern, not the spacing.

Step-by-step reasoning

1. (a) Screen moved away from the plane of the slits

When you increase $D$ , the fringe width $\beta = \frac{\lambda D}{d}$ increases proportionally. The fringes become broader and more spread out. The central maximum remains at the same angular position (straight ahead), but the linear distance to the first bright fringe on the screen becomes larger.

Why does this happen? The path difference between waves from $S_1$ and $S_2$ at a given point on the screen depends on the angle $\theta$ from the central line. For small angles, $\theta \approx \tan\theta = y/D$ , where $y$ is the distance from the centre. The condition for a bright fringe is $d \sin\theta = n\lambda$ , which becomes $d (y/D) = n\lambda$ , so $y = n\lambda D / d$ . Larger $D$ means larger $y$ for the same $n$ — the fringes literally stretch out.

Watch out

A common mistake is to think the fringes become fainter because the screen is farther. Yes, intensity drops with distance, but the question asks about the interference fringes — their spacing and visibility. The fringe width increases; the pattern is simply scaled up.

Effect: Fringe width increases. The pattern becomes more spread out.

2. (b) Separation between slits is increased

Here $d$ increases, so $\beta = \frac{\lambda D}{d}$ decreases. The fringes become narrower and more closely packed. The central maximum stays at the same place, but the first bright fringe moves closer to the centre.

Physically, a larger slit separation means that for a given angle $\theta$ , the path difference $d\sin\theta$ is larger. So the condition $d\sin\theta = n\lambda$ is satisfied at a smaller angle. The fringes crowd together.

Tip

If you double $d$ , the fringe width halves. This is a quick check: if you can’t see fringes because they are too fine, you might need to reduce $d$ .

Effect: Fringe width decreases. The fringes become finer and more numerous per unit length on the screen.

3. (c) The source slit is moved closer to the plane of the double slit

This is the subtle one. The source slit (the single slit before $S_1$ and $S_2$ ) is usually placed some distance away so that light spreads out and illuminates both slits equally. Moving it closer to the double slit changes the illumination. …

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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 29 on this concept.

CBSE 2026Set 55/1/11 markMCQ
$Q. The shape of the interference fringes in Young's double-slit experiment, when the distance between the slit and the screen is very large as compared to the slit-separation, is nearly (A) straight (B) parabolic (C) circular (D) hyperbolic$
›Reveal solutionSolution

When the screen is very far from the slits compared to their separation, points of constant path difference lie on nearly straight lines parallel to the slits, making the fringes straight.

The key to understanding fringe shape lies in recognizing what an interference fringe actually represents: it is the locus of all points on the screen where the path difference from the two slits is constant.

In Young's double-slit experiment, we have two coherent sources $S_1$ and $S_2$ separated by distance $d$ . A point $P$ on the screen at distance $D$ from the slits will show constructive or destructive interference depending on the path difference $\Delta = |S_2P - S_1P|$ .

For a bright fringe of order $n$ , we need $\Delta = n\lambda$ . The question is: what is the shape of the curve connecting all points $P$ that satisfy this condition?

Geometry of path difference

Consider a point $P$ on the screen at coordinates $(x, y)$ if we place the origin midway between the slits. The two slits are at positions roughly $(0, \pm d/2, 0)$ in 3D space, and the screen is at distance $D$ along the perpendicular.

The exact path difference is:
$\Delta = \sqrt{D^2 + (y - d/2)^2 + x^2} - \sqrt{D^2 + (y + d/2)^2 + x^2}$
This is the general equation for a hyperbola in the $xy$ -plane. So strictly speaking, fringes are hyperbolic curves.

The far-field approximation

Now comes the crucial condition: $D \gg d$ (screen distance much larger than slit separation).

When $D$ is very large, we can use the binomial approximation. For the path from $S_2$ to $P$ :
$S_2P = D\sqrt{1 + \frac{(y-d/2)^2 + x^2}{D^2}} \approx D + \frac{(y-d/2)^2 + x^2}{2D}$
Similarly for $S_1P$ . The path difference becomes:
$\Delta \approx \frac{(y-d/2)^2 - (y+d/2)^2}{2D} = \frac{-2yd}{2D} = -\frac{yd}{D}$
(The $x^2$ terms cancel out.)

For constant path difference $\Delta = n\lambda$ , we get:

$y = -\frac{n\lambda D}{d} = \text{constant}$ …
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CBSE 2026Set 55/2/11 markMCQ
$Q. Two statements are given - one labelled Assertion (A) and the other Reason (R). Select the correct answer from the codes below: (A) Both (A) and (R) are true and (R) is the correct explanation of (A). (B) Both (A) and (R) are true, but (R) is not the correct explanation of (A). (C) (A) is true, but (R) is false. (D) (A) is false and (R) is also false. Assertion (A) : Light added to light can produce darkness. Reason (R) : When two coherent light waves interfere, there is darkness at the position of destructive interference.$
›Reveal solutionSolution
Interference of coherent light waves can indeed produce darkness at points of destructive interference; both statements are true and the reason correctly explains the assertion. The answer is (A).

Why light can produce darkness

The assertion sounds paradoxical at first—how can adding light to light create darkness? The resolution lies in understanding that light is a wave phenomenon, and waves don't simply add arithmetically in intensity. Instead, they superpose according to their phase relationship.

When two coherent light waves (waves with a constant phase relationship) meet, their electric field amplitudes add vectorially. If they arrive in phase, the amplitudes reinforce (constructive interference, brighter light). If they arrive exactly out of phase—crest meeting trough—the amplitudes cancel (destructive interference), and the resultant intensity becomes zero or near-zero. This is darkness produced by adding light to light.

The reason statement captures precisely this mechanism: destructive interference between coherent waves creates regions of darkness.

Examining the statements

Assertion (A): "Light added to light can produce darkness"

This is experimentally verified in phenomena like Young's double-slit experiment, thin-film interference, and Newton's rings. At certain positions on the screen or observation plane, the intensity drops to zero despite light arriving from two sources. The statement is true.

Reason (R): "When two coherent light waves interfere, there is darkness at the position of destructive interference"

Destructive interference occurs when the path difference between two coherent waves is an odd multiple of half-wavelengths:

$\Delta = \left(m + \frac{1}{2}\right)\lambda, \quad m = 0, 1, 2, \ldots$ …
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CBSE 2026Set A1 markMCQ
$Q. The phase difference φ is related to path difference λ by (A) (λ/π)φ (B) (π/λ)φ (C) (λ/2π)φ (D) (2π/λ)φ$
›Reveal solutionSolution

Phase difference = (2π/λ) × path difference.

The fundamental relation between phase difference (Δϕ) and path difference (Δx) is

$\Delta\phi = \frac{2\pi}{\lambda}\,\Delta x.$

…

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CBSE 2026Set A1 markMCQ
$Q. For destructive interference, the path difference should be equal to (A) nλ (B) (2n+1)λ/2 (C) zero (D) infinity$
›Reveal solutionSolution

Destructive interference occurs when the path difference is an odd multiple of λ/2.

Two waves interfere destructively (cancel) when they arrive exactly out of phase, i.e. a phase difference of $\pi,3\pi,5\pi,\dots$ . In terms of path difference this means an odd multiple of half a wavelength:

…

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CBSE 2026Set ANNUAL1 mark
$Q. The displacement of water molecules at any instant on the surface of water at nodal lines is ______.$
›Reveal solutionSolution

Nodal lines are where two overlapping waves are always exactly out of phase, so their displacements cancel completely at every instant, leaving zero net displacement.

When two coherent water-wave sources produce overlapping ripples, at points on a nodal line the crest of one wave always coincides with the trough of the other (path difference = odd multiple of half wavelength), so d …

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CBSE 2025Set 55/5/11 markMCQ
$a$
›Reveal solutionSolution
When two coherent waves of equal amplitude $a$ interfere, the resultant amplitude varies from $0$ (destructive) to $2a$ (constructive); since intensity is proportional to the square of amplitude, the intensity range is 0 to $4a^2$ .

The heart of this problem lies in understanding how wave superposition affects intensity. When two coherent waves meet, they don't simply add their intensities — instead, their amplitudes add vectorially, and the resulting intensity depends on the square of this net amplitude.

For light waves, intensity $I$ is proportional to the square of the amplitude: $I \propto A^2$ . If we set the proportionality constant to unity for simplicity (which is standard when comparing relative intensities), then $I = A^2$ .

Now let's trace what happens when two coherent waves, each with amplitude $a$ , interfere.

The amplitude addition principle

At any point on the screen, the two waves arrive with some phase difference $\delta$ (which depends on the path difference). The resultant amplitude is found by vector addition:
$A_{\text{net}} = a_1 + a_2$
where $a_1$ and $a_2$ are the individual wave amplitudes treated as phasors. For two waves of equal amplitude $a$ with phase difference $\delta$ :
$A_{\text{net}} = \sqrt{a^2 + a^2 + 2a \cdot a \cos\delta} = a\sqrt{2(1 + \cos\delta)}$
Using the identity $1 + \cos\delta = 2\cos^2(\delta/2)$ :
$A_{\text{net}} = 2a\left|\cos\frac{\delta}{2}\right|$
Finding the intensity extremes

Maximum amplitude (constructive interference): When $\delta = 0, 2\pi, 4\pi, \ldots$ (waves in phase), we have $\cos(\delta/2) = 1$ , so:

$A_{\max} = 2a$
The maximum intensity is:

$I_{\max} = A_{\max}^2 = (2a)^2 = 4a^2$ …
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CBSE 2025Set 55/6/11 markMCQ
$I_0$
›Reveal solutionSolution
Interference redistributes light energy across the screen but cannot create or destroy it, so the average intensity equals the sum of the two individual intensities: $I_0 + I_0 = 2I_0$ . The correct option is (C).

When two coherent waves meet, they produce bright and dark fringes. At some points they add constructively (bright), at others destructively (dark). The question asks for the average intensity over the whole screen — not the maximum or minimum at any particular point.

The key insight is energy conservation. The two sources together deliver a fixed amount of energy to the screen. Interference only redistributes this energy — concentrating it in bright fringes and depleting it in dark ones — it cannot create or destroy energy. So the average over the pattern must equal what the two waves would deliver independently.

Let's confirm this with the intensity formula.

Write the resultant intensity at a point. For two coherent waves of intensity $I_0$ each, meeting with phase difference $\delta$ :

$I = I_0 + I_0 + 2\sqrt{I_0 \cdot I_0}\,\cos\delta = 2I_0(1 + \cos\delta)$
This varies from $I_{\text{max}} = 4I_0$ (at $\delta = 0, 2\pi, \ldots$ ) down to $I_{\text{min}} = 0$ (at $\delta = \pi, 3\pi, \ldots$ ).

Average over the pattern. As you move across the screen, the phase difference $\delta$ sweeps uniformly through all values from $0$ to $2\pi$ :

$\langle I \rangle = 2I_0\left(1 + \langle\cos\delta\rangle\right)$

Evaluate the average of cosine.

Over a complete cycle, $\langle\cos\delta\rangle = 0$ .

Conclude.

$\langle I \rangle = 2I_0(1 + 0) = 2I_0$ …
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CBSE 2025Set X11 markMCQ
$Q. Which one of the following statements is WRONG about interference of light?$
(a) Light waves of same wavelength coming from two independent sources can be coherent and can produce interference
(b) When the path difference between two interfering waves in $n\lambda$ , bright fringe is produced (Here $n = 0, 1, 2, \ldots$ and $\lambda$ is the wavelength of light)
(c) When the phase difference between two interfering waves is $(2n+1)\pi$ , dark fringe is produced (Here $n = 0, 1, 2, \ldots$ )
(d) In Young's double slit experiment, dark and bright fringes are equally spaced
›Reveal solutionSolution

(a) — this is the WRONG statement. Two independent sources cannot maintain a constant phase relationship, so they are not coherent and cannot produce a stable interference pattern; coherent sour …

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CBSE 2025Set D1 markMCQ
$Q. Two light waves of equal amplitude and equal wavelengths are superimposed. The amplitude of the resultant wave will be maximum when phase difference between them is (A) zero (B) π/4 (C) π/2 (D) π$
›Reveal solutionSolution

Two equal-amplitude waves add to a maximum when they are in phase, i.e. phase difference = 0.

For two waves of amplitude a with phase difference φ, the resultant amplitude is

A = √(a² + a² + 2a·a·cosφ) = 2a·cos(φ/2)

…

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CBSE 2025Set D1 markMCQ
$Q.The bubble of soap appears coloured due to (A) diffraction (B) polarization (C) interference (D) reflection$
›Reveal solutionSolution

Thin-film interference between the two reflected rays gives colour, since which wavelength interferes constructively depends on the film thickness.

A soap film is a very thin layer of liquid. Light reflected from its front surface interferes with light reflected from its back surface (thin-film interference).

…

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CBSE 2025Set A1 mark
$Q. Fill in the blank with appropriate word: ______ of light shows redistribution of energy.$
›Reveal solutionSolution

Interference of light merely redistributes energy between bright and dark regions; it does not create or destroy energy.

When two coherent light waves superpose, they produce a pattern of alternating bright and dark fringes. At points of constructive interference, the resultant intensity is greater than the sum of individual intensities (extra brightness), while at points of destructive interference the resultant intensity is less (often close to zero). Averaged over the whole pattern, however, the total energy is exactly conserved — …

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CBSE 2025Set ANNUAL1 markMCQ
$Q. Two waves whose intensity ratio is 9:1 produce interference. The ratio of maximum and minimum intensities will be$
(a) 10:8
(b) 9:1
(c) 4:1
(d) 2:1
›Reveal solutionSolution

In interference, I_max and I_min depend on the square root of the individual intensities, so a 9:1 intensity ratio does NOT give a 9:1 max:min ratio - it gives 4:1.

For two coherent sources of intensities I1 and I2, the resultant intensity varies between:

I_max = (sqrt(I1) + sqrt(I2))^2

I_min = (sqrt(I1) - sqrt(I2))^2

…

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Q.What is the effect on the interference fringes in Young's double slit experiment due to each of the following operations ? Justify your answers.

The core idea

Step-by-step reasoning

1. (a) Screen moved away from the plane of the slits

2. (b) Separation between slits is increased

3. (c) The source slit is moved closer to the plane of the double slit

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