Q.In a two-slit interference arrangement the slits $S_1$ and $S_2$ are separated by a distance $2D$; their midpoint is $C$ so that $S_1C = CS_2 = D$. A screen is placed parallel to the slit plane at a perpendicular distance $CO = D$ from the slits, where $O$ is the point on the screen directly opposite $C$. Hence the slit-to-screen distance ($D$) equals half the slit separation ($2D$). Monochromatic light of wavelength $\lambda$ is used. Obtain the value of $D$ in terms of $\lambda$ such that the first minimum on the screen falls at a point $P$ for which the distance $OP$ from the centre $O$ is equal to $D$.

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.

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:

The spacing between consecutive bright bands (fringe width) is:

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 …