(a) (i) In diffraction due to a single slit, the phase difference between light waves reaching a point on the

Question

Figure — Figure — CBSE 2023 55/4/1 Q29 — FigureFigure — CBSE 2023 55/4/1 Q29

Q. (a)

(i) In diffraction due to a single slit, the phase difference between light waves reaching a point on the screen is

5\pi

. Explain whether a bright or a dark fringe will be formed at the point.

(ii) What should the width (

a

) of each slit be to obtain eight maxima of two double-slit patterns (slit separation

d

) within the central maximum of the single slit pattern ?

(iii) Draw the plot of intensity distribution in a diffraction pattern due to a single slit.

(OR)

(b)

(i) In a Young's double-slit experiment

SS_2 - SS_1 = \dfrac{\lambda}{4}

, where

S_1

and

S_2

are the two slits as shown in the figure. Find the path difference

(S_2P - S_1P)

for constructive and destructive interference at P.

(ii) What is the effect on the interference fringes in a double-slit experiment, if the monochromatic source S is replaced by a source of white light ?

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Part (a)Concept understanding — Single Slit Diffraction

Single Slit Diffraction: From Intuition to Precision

Imagine you're standing at the edge of a swimming pool and you send a straight wave toward a narrow gap in a wall. If the gap is wide, the wave mostly goes straight through — a clean "shadow" behind the wall. But if the gap is tiny, something strange happens: the wave spreads out in all directions beyond the gap, like ripples from a pebble. That spreading is diffraction.

Light does the same thing. When a parallel beam of light passes through a single narrow slit, it doesn't just make a sharp rectangle on a screen. Instead, you get a pattern: a bright central band, then dark bands (minima), then weaker bright bands (maxima), alternating as you move outward. The narrower the slit, the more the light spreads.

Why does this happen? The core idea

Light from every point across the slit travels to the screen. At any point on the screen, the light arriving from different parts of the slit has travelled different distances. If those path differences are exactly half a wavelength ( $\lambda/2$ ), the waves cancel — you get darkness. If they are a whole wavelength ( $\lambda$ ), they reinforce — you get a weaker bright band.

The key is that the slit is not a point source. It's a continuous line of sources, each sending out Huygens wavelets. The pattern is the result of interference among all those wavelets.

Watch out

Common mistake

Students often think diffraction is just "bending around corners." That's part of it, but the real physics is interference between wavelets from different parts of the same slit. Without that interference, there would be no alternating dark and bright bands — just a fuzzy blur.

The precise condition for minima

Let the slit width be $a$ and the wavelength be $\lambda$ . For a point on a screen far away (the Fraunhofer or far-field condition), light rays from the slit are nearly parallel. The path difference between a wavelet from the top edge and one from the centre is $\frac{a}{2} \sin\theta$ , where $\theta$ is the angle from the straight-through direction.

For the first minimum, the wavelets from the top half of the slit cancel those from the bottom half exactly. That happens when the path difference between the two edges is exactly one wavelength:

a \sin\theta = \lambda

For the second minimum, the slit can be divided into four equal zones, each cancelling the next, giving:

a \sin\theta = 2\lambda

In general, the condition for dark fringes (minima) is:

a \sin\theta = m\lambda \quad \text{for } m = \pm 1, \pm 2, \pm 3, \dots

Notice $m = 0$ is not a minimum — it's the centre of the bright central maximum.

What about the maxima?

The maxima occur roughly halfway between minima, but their positions are not given by a simple formula like $a \sin\theta = (m + \frac12)\lambda$ . That formula works for double-slit interference, but for a single slit the maxima are slightly shifted. The exact positions come from solving a calculus problem (the derivative of the intensity function), but for exams you only need the minima condition and the fact that the central maximum is twice as wide as the others.

Tip

Quick exam fact

The angular width of the central maximum is $2\theta_1$ , where $\theta_1$ satisfies $a \sin\theta_1 = \lambda$ . So the central maximum spans from $-\lambda/a$ to $+\lambda/a$ in $\sin\theta$ .

The intensity pattern (qualitative) …

Part (b)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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Part (a): a phase difference of $5\pi$ is an odd multiple of $\pi$ $\Rightarrow$ a bright secondary maximum; eight interference maxima fit the central diffraction maximum when $a=d/4$ ; the single-slit intensity is $I_0(\sin\beta/\beta)^2$ .

Part (b): the $\lambda/4$ source-side path shifts the pattern, so constructive at $S_2P-S_1P=n\lambda-\tfrac{\lambda}{4}$ and destructive at $(2n+1)\tfrac{\lambda}{2}-\tfrac{\lambda}{4}$ ; white light gives a white central fringe with coloured fringes that fade to white.

Diagram (a) of a plane wave diffracting through a narrow single slit and (b) the resulting intensity distribution on a viewing screen, showing a bright central band flanked by progressively fainter secondary bright and dark fringes.

Part (a) — Single-slit diffraction

(i) Phase difference $5\pi$ . Divide the slit into strips; the resultant amplitude is

A=A_0\frac{\sin\beta}{\beta},\qquad \beta=\frac{\phi}{2}=\frac{\pi a\sin\theta}{\lambda},

where $\phi$ is the phase difference between the extreme rays. Minima require $\sin\beta=0$ with $\beta\neq0$ , i.e. $\beta=m\pi\Rightarrow\phi=2m\pi$ . Secondary maxima lie close to $\beta=(m+\tfrac12)\pi\Rightarrow\phi=(2m+1)\pi$ .

For $\phi=5\pi$ , $\beta=2.5\pi$ , $\sin\beta=\sin(2.5\pi)=+1$ , so $A=A_0/2.5\pi\neq0$ . Since $5\pi$ is an odd multiple of $\pi$ (not a multiple of $2\pi$ ), the point is a secondary maximum — a bright fringe.

Watch out

Single-slit minima are at phase difference $2\pi,4\pi,\dots$ (path $=m\lambda$ ), not at odd multiples of $\pi$ . So $5\pi$ is not a minimum.

(ii) Eight maxima inside the central maximum. The central diffraction maximum reaches the first diffraction minimum at $\sin\theta=\lambda/a$ . Interference maxima occur at $d\sin\theta=n\lambda$ . Setting the 4th-order interference maximum at the diffraction-minimum edge,

\frac{4\lambda}{d}=\frac{\lambda}{a}\;\Rightarrow\; \boxed{a=\frac{d}{4}}.

This places four interference maxima on each side of the centre within the central diffraction envelope.

(iii) Intensity plot.

$I=I_0\left(\frac{\sin\beta}{\beta}\right)^2,\qquad\beta=\frac{\pi a\sin\theta}{\lambda}.$ …

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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 58 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 V11 mark
$Q.The bending of light around the corners and entering into the geometric shadow region is called __________. Fill in the blank choosing the appropriate answer from the bracket: (photons, diffraction, polarity, monopoles, greater than unity, less than unity)$
›Reveal solutionSolution

diffraction …

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CBSE 2026Set ANNUAL1 mark
$Q.Read the following passage carefully and answer the questions given below - Light entering in a dark room through a narrow gap under or around a closed door often appears to bend and spread. This is due to diffraction, the narrow gap acts like a slit, and light wave bends around the edges. This phenomenon is a small but clear demonstration of diffraction.$
(a) What would be the approximate size of sharp edge or opening compared to the wavelength of light for diffraction to be clearly observed?
›Reveal solutionSolution

Diffraction is prominent only when the aperture/obstacle size is comparable to the wavelength of light.

Diffraction effects become clearly noticeable only when the size of the slit/gap or obstacle is of the same order of magnitude as (comparable to) the wavelength of light used. If the opening is much larger than the wavelength, the bending is negligible and light appears to travel in straight lines (ray optics applies); as the opening size approaches the wavelength, diffraction (bending and spreading) becomes prominent — which is why the narrow …

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CBSE 2026Set ANNUAL1 mark
$Q.Read the following passage carefully and answer the questions given below - Light entering in a dark room through a narrow gap under or around a closed door often appears to bend and spread. This is due to diffraction, the narrow gap acts like a slit, and light wave bends around the edges. This phenomenon is a small but clear demonstration of diffraction.$
(b) Why can the diffraction not be explained by ray-optics?
›Reveal solutionSolution

Ray optics assumes light always travels in straight lines and cannot describe the wave interference behind diffraction.

Ray (geometrical) optics treats light purely as straight-line rays and predicts that an obstacle or slit will simply produce sharp-edged shadows/beams, with no bending at the edges. Diffraction, however, is fundamentally a wave phenomenon — it arises from the superposition (interference) of secondary wavelets originating from different points of the same wavefront as it passes an edge or narrow opening (Huygens–Fresnel principle). Since ray optics ignores the wave nature of …

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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 2026Set ANNUAL1 mark
$Q.Define diffraction of light.$
›Reveal solutionSolution

Diffraction is the deviation of light from a straight-line path when it passes an obstacle or a narrow opening whose size is comparable to its wavelength.

Diffraction of light is the phenomenon of bending of light waves around the corners of an obstacle or spreading of light after passing through a narrow slit/aperture, so that light appears in regions that would be a dark geometrical shadow according to simple ray optics. It becomes noticeable when the size of the obstacle/aperture is comparable to the wavelength of light, and is a direct consequence of the wave nature of light (a manifestatio …

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CBSE 2026Set ANNUAL1 markMCQ
$Q.The size of the obstacle for the diffraction of light should be$
(a) much larger than the wavelength of light
(b) much smaller than the wavelength of light
(c) of the order of wavelength of light
(d) anything can happen
›Reveal solutionSolution

Diffraction (bending of waves around obstacles) is significant only when the obstacle or slit size is comparable to the wavelength - much bigger or much smaller sizes don't show it clearly.

Diffraction is the bending/spreading of waves as they pass an obstacle or through an aperture. This spreading is only prominent when the size of the obstacle/aperture (a) is of the SAME ORDER as the wavelength (lambda) of the wave, i.e. a ~ lambda. If the obstacle is much LARGER than the wavelength, the wave essentially travels in straight lines (geometrical shadow, negligible diffraction) - this is why we don't see visible light (wavelength ~500 nm) diffracting around everyday-s …

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CBSE 2026Set ANNUAL1 mark
$Q.What do you understand by the term 'diffraction of light'?$
›Reveal solutionSolution

Diffraction is the bending of light into the region that geometrical (ray) optics would call 'shadow' when it passes an obstacle or aperture.

According to simple ray/geometrical optics, light travelling past an obstacle or through a slit should produce a sharp-edged shadow. In reality, being a wave, light bends slightly around the edges of the obstacle/aperture and spreads into what would otherwise be the shadow region, producing a pattern of bright and dark fringes near the edges. This bending and spreading of light waves around obstacles/apertures is called diffraction. It becomes prominent only when the size of the obstacle or aperture is comparable to the wavelength of light, …

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CBSE 2026Set ANNUAL1 markMCQ
$Q. Select the correct option with respect to the figures given below:$
(a) Fig.
(i) depicts diffraction pattern and Fig.
(ii) depicts interference pattern
(b) Both depict interference pattern
(c) Both depict diffraction pattern
(d) Fig.
(i) depicts an interference pattern due to a double-slit and Fig.
(ii) depicts a diffraction pattern due to a single-slit
›Reveal solutionSolution
The key distinguishing feature between interference and diffraction intensity patterns is the relative heights of the fringes: interference from two narrow slits gives many fringes of roughly equal intensity, while diffraction from a single slit gives one strong central maximum with much weaker, rapidly falling secondary maxima.

Distinguishing interference from diffraction patterns

In Young's double-slit interference experiment, two coherent narrow sources produce a pattern of bright and dark fringes on the screen. Because the two slits are treated as (nearly) point/line sources of equal amplitude, the resulting bright fringes are all of comparable/roughly equal intensity across the region observed (there is a slowly-varying diffraction "envelope" from each slit's own finite width, but for the idealised interference pattern, the fringes near the centre appear as a series of similar-height peaks, evenly spaced by the fringe width $\beta=\lambda D/d$ ).

In single-slit diffraction, light passing through one slit produces a pattern with:

one very bright central maximum (roughly twice as wide as the secondary maxima), and

a series of much fainter secondary maxima on either side, whose intensities fall off rapidly (roughly as $1/m^2$ or faster) with distance from the centre.

Applying this to the two figures

…
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Q.(a)

Part (a) — Single-slit diffraction

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