(a) What are coherent sources? Why are they necessary for observing a stable interference pattern? Draw a grap

Question

Q. (a) What are coherent sources? Why are they necessary for observing a stable interference pattern? Draw a graph showing the variation of intensity of light with the position on the screen in Young's double-slit experiment.

(b) Find the intensity of light at a point on the screen when two interfering waves of the same intensity (

I_0

) have a path difference of

(i)

\dfrac{\lambda}{4}

and

(ii)

\dfrac{\lambda}{3}

(OR)

(a) Draw a labelled ray diagram of a refracting telescope when it forms the image of a distant object at infinity. Derive an expression for its magnifying power.

(b)

(i) In a telescope the objective has a much larger aperture than the eyepiece. Why?

(ii) Write two advantages of a reflecting telescope over a refracting telescope.

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Part (a)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:

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

Part (b)Concept understanding — Angular Magnification

What is Angular Magnification?

When you look at a tiny object — say a grain of salt — you hold it close to your eye to see it bigger. But there is a limit: bring it too close and it blurs. The closest distance at which your eye can focus comfortably is called the near point, conventionally taken as 25 cm for a normal eye. At that distance, the object subtends a certain angle at your eye. That angle determines how large it appears — not its physical size, but the fraction of your field of view it occupies.

Now imagine using a magnifying glass. The same grain of salt now looks much larger. Why? Because the lens lets you bring the object even closer than 25 cm while still seeing a clear, magnified image. That image is formed at a comfortable viewing distance, but the angle it subtends at your eye is far bigger than the angle the object would subtend at 25 cm without the lens.

Angular magnification is simply the ratio of these two angles:

Angular magnification $M = \frac{\theta_{\text{image}}}{\theta_{\text{object}}}$

where $\theta_{\text{image}}$ is the angle subtended by the image when viewed through the instrument, and $\theta_{\text{object}}$ is the angle subtended by the object when viewed with the naked eye at the near point (25 cm).

Why "Angular" and Not "Linear"?

A common confusion: a microscope or telescope does not give you a physically larger object — it gives you a larger apparent size. The image on your retina is bigger because the rays entering your eye are steeper. That steepness is measured by the angle. So magnification here is about angles, not actual lengths.

Important

Angular magnification is dimensionless. It tells you how many times wider the image appears compared to the object seen directly at the near point.

A Concrete Example

Take a simple magnifier (a convex lens) of focal length $f = 5$ cm. You place the object just inside the focal point so that a virtual, erect image forms at infinity (or at the near point). For the "image at infinity" case, the angle subtended by the image is $\theta_{\text{image}} \approx h / f$ , where $h$ is the object height. The angle subtended by the object at the near point (25 cm) is $\theta_{\text{object}} \approx h / 25$ .

Thus:

M = \frac{h/f}{h/25} = \frac{25}{f}

For $f = 5$ cm, $M = 5$ . The image appears 5 times larger than the object seen at 25 cm.

Tip

For a magnifier, the formula $M = 1 + \frac{25}{f}$ applies when the image is formed at the near point (25 cm) — giving slightly higher magnification than the infinity-focus case.

The Big Picture

Angular magnification is the language of all optical instruments:

Simple magnifier: $M \approx 25/f$ (image at infinity) …

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Part (a): coherent sources (same frequency, fixed phase difference) are needed for stable fringes; with $I=2I_0(1+\cos\phi)$ , a path difference $\lambda/4$ gives $2I_0$ and $\lambda/3$ gives $I_0$ . Part (b): a refracting telescope in normal adjustment has magnifying power $m=f_o/f_e$ ; a large objective collects more light and resolves better, and reflectors beat refractors on chromatic aberration and large-aperture construction.

Graph of intensity I versus position x on the screen for Young's double-slit interference, showing a series of equally spaced, equal-height cos-squared bright fringes of peak intensity 4I0 with dark minima between them.

Part (a)

Coherent sources are sources that emit light waves of the same frequency maintaining a constant phase difference over time. For a stable (time-averaged) interference pattern the bright and dark fringes must stay at fixed positions; this only happens if the phase difference between the two waves does not change. Independent sources have phases that fluctuate randomly (every $\sim10^{-8}$ s), so their pattern shifts too fast to see and the intensity everywhere averages to a uniform value. In Young's double-slit experiment coherence is obtained by deriving both slits from a single wavefront.

The intensity on the screen follows $I=I_{\max}\cos^2\!\left(\dfrac{\pi\Delta}{\lambda}\right)$ with $I_{\max}=4I_0$ : a central maximum of $4I_0$ , minima (zero) at $\Delta=\pm\lambda/2,\pm3\lambda/2,\dots$ , and equal secondary maxima of $4I_0$ at $\Delta=\pm\lambda,\pm2\lambda,\dots$ (a $\cos^2$ curve versus screen position).

I=I_1+I_2+2\sqrt{I_1I_2}\cos\phi = 2I_0(1+\cos\phi),\qquad \phi=\frac{2\pi}{\lambda}\,\Delta.

$\Delta=\dfrac{\lambda}{4}\Rightarrow \phi=\dfrac{2\pi}{\lambda}\cdot\dfrac{\lambda}{4}=\dfrac{\pi}{2}$ : $I=2I_0\left(1+\cos\tfrac{\pi}{2}\right)=2I_0(1+0)=2I_0.$
$\Delta=\dfrac{\lambda}{3}\Rightarrow \phi=\dfrac{2\pi}{3}$ : …

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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 36 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 55/3/11 markMCQ
$144$
›Reveal solutionSolution
For a telescope in normal adjustment, the magnifying power is the ratio of objective to eyepiece focal lengths ( $M = f_o / f_e$ ), and the tube length is their sum ( $L = f_o + f_e$ ). Here, $M = 144/6 = 24$ and $L = 144 + 6 = 150$ cm, so the correct option is (A).

The question gives you a telescope with an objective of focal length $f_o = 144$ cm and an eyepiece of focal length $f_e = 6.0$ cm. You need the magnifying power and the length of the telescope tube.

The key idea is that a telescope is used to view distant objects. Light from a faraway object arrives as nearly parallel rays. The objective lens forms a real, inverted image at its focal plane. The eyepiece then acts as a magnifier to view that image.

For relaxed viewing (normal adjustment), the eyepiece is adjusted so that the final image is at infinity. This means the image formed by the objective must lie exactly at the focal point of the eyepiece. So the distance between the two lenses — the tube length — is simply the sum of their focal lengths.

The magnifying power (or angular magnification) is defined as the ratio of the angle subtended at the eye by the final image to the angle subtended by the object when viewed directly. For a telescope in normal adjustment, this ratio simplifies beautifully to the ratio of the focal lengths.
$M = \frac{f_o}{f_e} \quad \text{and} \quad L = f_o + f_e$
Let's apply this directly.

Magnifying power:

$M = \frac{f_o}{f_e} = \frac{144}{6.0} = 24$

Tube length:

$L = f_o + f_e = 144 + 6.0 = 150 \text{ cm}$
So the magnifying power is 24 and the tube length is 150 cm. …
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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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Q.(a) What are coherent sources? Why are they necessary for observing a stable interference pattern? Draw a graph showing the variation of intensity of light with the position on the screen in Young's double-slit experiment.

Part (a)

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