(a) (i) A ray of light passes through a triangular prism. Show graphically how the angle of deviation varies w

Question

Q. (a)

(i) A ray of light passes through a triangular prism. Show graphically how the angle of deviation varies with the angle of incidence. Hence define the angle of minimum deviation.

(ii) A ray of light is incident normally on a refracting face of a prism of prism angle

A

and suffers a deviation of angle

\delta

. Prove that the refractive index

n

of the material of the prism is given by

n = \dfrac{\sin(A+\delta)}{\sin A}

(iii) The refractive index of the material of a prism is

\sqrt{2}

. If the refracting angle of the prism is

60^\circ

, find the

(1) angle of minimum deviation, and

(2) angle of incidence.

(OR)

(b)

(i) State Huygens' principle. A plane wave is incident at an angle

i

on a reflecting surface. Construct the corresponding reflected wavefront. Using this diagram, prove that the angle of reflection is equal to the angle of incidence.

(ii) What are coherent sources of light ? Can two independent sodium lamps act like coherent sources ? Explain.

(iii) A beam of light consisting of a known wavelength

520\ \text{nm}

and an unknown wavelength

\lambda

, in a Young's double-slit experiment, produces two interference patterns such that the fourth bright fringe of the unknown wavelength coincides with the fifth bright fringe of the known wavelength. Find the value of

\lambda

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Part (a)Concept understanding — Angle of Minimum Deviation and Refractive Index of Prism

Plotting deviation $d$ against angle of incidence $i_1$ for a prism gives a curve that falls to a single minimum value $D$ (the angle of minimum deviation) before rising again; at this minimum the ray path through the prism is symmetric, with the incident and emergent angles equal ( $i_1=i_2=i$ ), the two internal refraction angles equal ( $r_1=r_2=r=A/2$ , since $r_1+r_2=A$ ), and the ray inside the prism running parallel to its base. Substituting $i_1=i_2=i$ , $r_1=r_2=A/2$ and $i=(A+D)/2$ (from $D=2i-A$ ) into Snell's law $n=\sin i/\sin …

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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Figure — Part (a)(i) asks to show graphically how the angle of deviation varies with the angle of incidence to define m

Part (a): for a prism $n=\dfrac{\sin(A+\delta)}{\sin A}$ ; with $n=\sqrt2$ , $A=60^\circ$ gives $\delta_m=30^\circ$ , $i=45^\circ$ .

Part (b): Huygens gives $\angle r=\angle i$ ; independent lamps are incoherent; $\lambda=650\ \text{nm}$ .

Part (a)

(i) As the angle of incidence $i$ increases, the deviation $\delta$ first decreases, reaches a single minimum $\delta_m$ , then increases; the graph is a smooth curve concave upward. The angle of minimum deviation is the smallest value of $\delta$ , and it occurs when the ray passes symmetrically ( $r_1=r_2$ , $i=e$ ).

(ii) At normal incidence on the first face, $i_1=0$ so the ray enters undeviated; the whole prism angle appears at the second face. Refraction at the second face gives a total deviation $\delta$ , and geometry gives $r=A$ inside, with emergence angle $e$ where

\delta=e-A,\qquad n=\frac{\sin e}{\sin A}=\frac{\sin(A+\delta)}{\sin A}.

(iii) Using $n=\dfrac{\sin\frac{A+\delta_m}{2}}{\sin\frac{A}{2}}$ with $n=\sqrt2,\ A=60^\circ$ :

\sin\frac{60^\circ+\delta_m}{2}=\sqrt2\,\sin30^\circ=\sqrt2\times\tfrac12=\tfrac{1}{\sqrt2}=\sin45^\circ.

\frac{60^\circ+\delta_m}{2}=45^\circ\Rightarrow \delta_m=30^\circ.

Angle of incidence at minimum deviation: …

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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 42 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 markMCQ
$Q. The angle of minimum deviation of a prism depends on :$
(i) refractive index of the material of the prism
(ii) refractive index of surrounding medium
(iii) refracting angle of the prism
(a) Only option
(i)
(b) Only option
(ii)
(c) Only option
(iii)
(d) All (i),
(ii) and (iii)
›Reveal solutionSolution

(d) All (i), (ii) and (iii) …

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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 markMCQ
$Q. The graph between angle of incidence$
(i) and angle of deviation (δ) for a triangular prism is
(a) graph: monotonically decreasing curve, flattening out (shown in the original paper)
(b) graph: straight line increasing linearly from origin (shown in the original paper)
(c) graph: U-shaped curve with a minimum (shown in the original paper)
(d) graph: straight line decreasing linearly (shown in the original paper)
›Reveal solutionSolution

The angle-of-deviation vs angle-of-incidence graph for a prism is U-shaped: deviation falls to a minimum value (the angle of minimum deviation, where i = e) and then rises again.

As the angle of incidence i on a prism is increased from a small value, the deviation delta first decreases, reaches a single minimum value delta_m at the position of symmetric passage through the prism (where the angle of incidence equals the angle of emergence, i = e), and then increases again as i is increased further. This non-monoton …

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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
$39.5^\circ$
›Reveal solutionSolution

The prism formula relates the refractive index to the angle of minimum deviation and the prism angle; substituting $A=60^\circ$ (equilateral) and $D_m=39.5^\circ$ gives $\mu\approx1.53$ .

The prism (minimum-deviation) formula

For light passing through a prism of angle $A$ at the angle of minimum deviation $D_m$ , the refractive index of the prism material is

$\mu=\frac{\sin\!\left(\dfrac{A+D_m}{2}\right)}{\sin\!\left(\dfrac{A}{2}\right)}$

Given values

Since the prism is equilateral, $A=60^\circ$ . We are given $D_m=39.5^\circ$ .

…

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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 IMPROVEMENT1 markMCQ
$Q.What is the angle of minimum deviation for a prism made of a material with refractive index n and a very small prism angle A?$
(a) (n+1)A
(b) (n-1)A
(c) $\dfrac{(n+1)}{(n-1)}A$
(d) $\dfrac{(n-1)}{(n+1)}A$
›Reveal solutionSolution

For a thin prism (small angle A), the minimum deviation is $\delta_m = (n-1)A$ .

For a prism of refracting angle $A$ and refractive index $n$ , the general relation for minimum deviation is $n = \dfrac{\sin\left(\frac{A+\delta_m}{2}\right)}{\sin\frac{A}{2}}$ . When the prism angle $A$ is very small, the angle of minimum deviation $\delta_m$ is also small, so the sines can be replaced by the angles themselves ( …

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Q.(a)

Part (a)

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