(a) In Young's double-slit experiment, find the resultant intensity at points at which the interfering waves o

Question

I_0

(i)

\dfrac{\lambda}{3}

, and

(ii)

\dfrac{\lambda}{2}

(OR)

(b) A point source of light in air is kept at a distance of 12 cm in front of a convex spherical surface of glass of refractive index 1.5 and radius of curvature 30 cm. Find the nature and position of the image formed.

CBSECBSE Class XII Board 2025Subjective· 2mImportance★★★★★

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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:

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

Part (a): with $I=4I_0\cos^2(\phi/2)$ , a path difference $\lambda/3$ gives $I=I_0$ and $\lambda/2$ gives $I=0$ . Part (b): refraction at the convex surface gives $v=-22.5$ cm — a virtual image 22.5 cm in front of the surface.

Intensity in Young's double-slit experiment

Two coherent waves from the slits superpose; the resultant is governed by the phase difference $\phi=\dfrac{2\pi}{\lambda}\Delta x$ . For equal intensities $I_0$ ,

I=4I_0\cos^2\!\left(\frac{\phi}{2}\right).

$\Delta x=\lambda/3$ : $\phi=\dfrac{2\pi}{\lambda}\cdot\dfrac{\lambda}{3}=\dfrac{2\pi}{3}$ , so

I=4I_0\cos^2\!\frac{\pi}{3}=4I_0\left(\frac12\right)^2=I_0.

$\Delta x=\lambda/2$ : $\phi=\pi$ , so

I=4I_0\cos^2\!\frac{\pi}{2}=0\quad(\text{destructive interference}).

Watch out

For coherent sources the answer is not $I_0+I_0=2I_0$ ; interference reshapes the sum between $0$ and $4I_0$ .

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Concept understanding — Refraction at a Spherical Surface

Refraction at a Spherical Surface

Imagine you are looking at a fish in a pond. The fish appears closer to the surface than it actually is. That is refraction — light bends when it moves from water into air. Now imagine the boundary between the two media is not flat, but curved like the surface of a lens or a glass marble. That is a spherical refracting surface.

The Intuition

When light hits a flat surface (like a glass slab), it bends once and travels straight. But when the surface is curved, something more interesting happens. Each ray of light strikes the curve at a slightly different angle, so each ray bends by a different amount. The result is that all rays leaving one point can be made to converge to (or diverge from) another point — forming an image.

Think of a spherical surface as a tiny piece of a sphere. The centre of that sphere is called the centre of curvature $C$ , and the distance from the surface to $C$ is the radius of curvature $R$ . The line joining the centre of curvature to the centre of the surface is the principal axis.

The Sign Convention

Before we write the formula, we need a consistent way to measure distances. The standard convention (Cartesian sign convention) is:

Distances measured against the direction of incident light are negative.
Distances measured along the direction of incident light are positive.
The pole $P$ (the vertex of the spherical surface) is the origin.

So for a convex surface (bulging toward the incident light), $R$ is positive. For a concave surface (curving away), $R$ is negative. Object distance $u$ is always negative (object is on the incident side). Image distance $v$ can be positive or negative depending on where the image forms.

The Derivation in One Paragraph

Consider a point object $O$ on the principal axis. A ray from $O$ strikes the spherical surface at point $A$ and bends according to Snell's law: $n_1 \sin i = n_2 \sin r$ . For small angles (paraxial approximation), $\sin\theta \approx \theta$ , so $n_1 i = n_2 r$ . Using geometry, $i = \alpha + \gamma$ and $r = \gamma - \beta$ , where $\alpha$ , $\beta$ , $\gamma$ are the angles the ray makes with the principal axis at $O$ , $I$ , and $C$ respectively. Substituting and using $\alpha \approx \tan\alpha = \frac{AP}{PO}$ , $\beta \approx \frac{AP}{PI}$ , $\gamma \approx \frac{AP}{PC}$ , and cancelling $AP$ , you get the relation.

\frac{n_2}{v} - \frac{n_1}{u} = \frac{n_2 - n_1}{R}

Here:

$n_1$ = refractive index of the medium where the object lies
$n_2$ = refractive index of the medium where the image forms
$u$ = object distance from pole (negative)
$v$ = image distance from pole (positive if real image on the opposite side)
$R$ = radius of curvature (positive if centre of curvature is on the image side)

What This Formula Tells You

The formula is a single equation that connects four things: where the object is, where the image forms, how curved the surface is, and what the two media are. If you know any three, you can find the fourth.

For a convex surface (say, light going from air into glass through a convex surface), $R > 0$ and $n_2 > n_1$ , so the right side is positive. This means $v$ is positive — the image forms on the other side. The surface converges light.

For a concave surface (light going from air into glass through a concave surface), $R < 0$ , so the right side is negative. The surface diverges light. …

Refraction at a convex spherical surface

Light from a point source in air (medium 1) refracts into glass (medium 2) through a convex surface:

\frac{n_2}{v}-\frac{n_1}{u}=\frac{n_2-n_1}{R}.

Assign signs (Cartesian convention, light travelling left to right): object in air 12 cm to the left, $u=-12\ \text{cm}$ ; convex surface toward the object, centre of curvature on the transmitted side, $R=+30\ \text{cm}$ ; $n_1=1$ , $n_2=1.5$ .
Substitute: …