JEE Mains Physics · Wave Optics

Double Slit Intensity and Slab Shifts

The brightness at any point on a double-slit screen is I_max cos²(φ/2), with the phase φ found from the path difference; a thin sheet over one slit adds (μ − 1)t to that path and slides the whole pattern towards the covered slit.

All Wave Optics notes JEE Mains Physics strategy

Why this matters

Twenty PYQs, nine of them multiple choice, and eight from 2026, more than any other page in the chapter. Five work out the path difference from the geometry, most of them at the point straight opposite one slit. Eight turn a path or phase difference into an intensity. Seven put a thin sheet over one slit and ask how far, or by how many fringes, the pattern moves.

Concept 1 of 3: Path difference from the geometry

Every intensity question starts with the path difference. For a point a height y up a distant screen, the two paths differ by yd/D. The point straight opposite one slit is a favourite, because there y is exactly d/2. When the screen is close, or the source is not on the centre line, write each path out in full instead.

Definition

Far screen (D much larger than d and y): $\Delta x = \dfrac{yd}{D}$ .
Straight opposite one slit: $y = \dfrac{d}{2}$ , so $\Delta x = \dfrac{d^{2}}{2D}$ .
The first dark fringe falls opposite a slit when $\dfrac{d^{2}}{2D} = \dfrac{\lambda}{2}$ , that is $\lambda = \dfrac{d^{2}}{D}$ .
Otherwise write each path with Pythagoras. A point P a distance D straight ahead of one slit is $\sqrt{D^{2} + d^{2}}$ from the other, so $\Delta x = \sqrt{D^{2} + d^{2}} - D \approx \dfrac{d^{2}}{2D}$ .
If the source is off the centre line, the paths from the source to the two slits differ too; add that difference to the one after the slits.

Path difference

\Delta x = \frac{yd}{D}, \qquad \Delta x_{\text{opposite a slit}} = \frac{d^{2}}{2D}, \qquad \sqrt{D^{2} + d^{2}} - D \approx \frac{d^{2}}{2D}

Worked example

Slits 0.8 mm apart are lit with light of 400 nm, and the screen is 1.6 m away. What is seen on the screen straight opposite one of the slits?

Practice this conceptself-check · 4 quick reps

The same idea in a real exam question:

JEE Mains · 2026 · 5 Apr 2026 Shift 2 · Q17Moderate

Example 1 · Wave Optics · Double Slit Intensity and Slab Shifts

The maximum intensity in a Young's double slit experiment is

I_{0}

. Distance between the slits (d) is

5\lambda

, where

\lambda

is the wavelength of light used. The intensity of the fringe, exactly opposite to one of the slits on the screen, placed at

D = 10\text{ }d

\_\_\_\_

Opposite a slit is y = d/2

The centre of the screen is midway between the slits, so a slit is d/2 from it, not d. Using y = d doubles the path difference.

yd/D needs a far screen

When the screen is only a few slit gaps away, or a figure sets the point beside one slit, write each path with Pythagoras and subtract.

Know which I₀ the question means

Some stems call the maximum intensity I₀, others call the intensity of one slit I₀, which makes the maximum 4I₀. Read the definition before you answer.

Drill 4 more on path difference from the geometry

Concept 2 of 3: Intensity from path and phase

With two equal slits, the brightness rises and falls smoothly as cos² of half the phase difference. At the centre the waves are in step and the screen is four times as bright as one slit alone. Half a wave of path later it is dark. Everything in between follows one formula.

Definition

Phase from path: $\phi = \dfrac{2\pi}{\lambda}\Delta x$ .
Equal slits, each giving $I_{0}$ alone: $I = 4I_{0}\cos^{2}\dfrac{\phi}{2} = I_{\max}\cos^{2}\dfrac{\phi}{2}$ , with $I_{\max} = 4I_{0}$ .
Equivalently $I = I_{\max}\cos^{2}\dfrac{\pi\,\Delta x}{\lambda}$ .
Two landmarks: $\Delta x = \lambda/4$ gives $I_{\max}/2$ ; $\Delta x = \lambda/2$ gives zero.
To find where a given intensity first appears: solve for the smallest $\phi$ , turn it into $\Delta x$ , then $y = \Delta x\,\dfrac{D}{d}$ .

Double-slit intensity

I = 4I_{0}\cos^{2}\frac{\phi}{2} = I_{\max}\cos^{2}\frac{\pi\,\Delta x}{\lambda}, \qquad \phi = \frac{2\pi}{\lambda}\Delta x

Worked example

Light of 640 nm falls on slits 0.3 mm apart, with the screen 1.5 m away. How far from the central maximum does the intensity first fall to half the maximum?

Practice this conceptself-check · 4 quick reps

The same idea in a real exam question:

JEE Mains · 2024 · 9 April 2024 · Q24Moderate

Example 2 · Wave Optics · Double Slit Intensity and Slab Shifts

In a Young's double slit experiment, the intensity at a point is

\left( \frac{1}{4} \right)^{\text{th~}}

of the maximum intensity, the minimum distance of the point from the central maximum is______

\mu m

. (Given:

\lambda= 600\text{ }nm,\text{ }d = 1.0\text{ }mm,D = 1.0\text{ }m

)

Half the phase inside the cosine

I = I_max cos²(φ/2), not cos²φ. With the full phase, λ/4 would give zero instead of half.

I₀ for one slit means I_max = 4I₀

If I₀ is the intensity of one slit alone, the formula is 4I₀cos²(φ/2). Writing I₀cos²(φ/2) gives answers four times too small.

Path is not phase

A path of λ/3 is a phase of 2π/3. Put the path straight into the cosine and the answer is wrong by a factor of 2π/λ.

Drill 7 more on intensity from path and phase

Concept 3 of 3: A thin sheet over one slit

A glass sheet over one slit makes that path longer by (μ − 1)t. The central bright fringe is wherever the two optical paths are equal, so it moves towards the covered slit, where the geometric path is shorter and makes up the difference. The whole pattern slides over; the fringes keep their width.

Definition

Extra optical path: $(\mu - 1)t$ . Number of fringes shifted: $N = \dfrac{(\mu - 1)t}{\lambda}$ .
Distance shifted: $\Delta y = N\beta = \dfrac{(\mu - 1)tD}{d}$ , towards the covered slit.
Two sheets of equal thickness, one on each slit: net extra path $(\mu_{2} - \mu_{1})t$ , shift towards the slit with the larger μ.
The fringe width does not change.
The point where the centre used to be is bright again only if $(\mu - 1)t$ is a whole number of wavelengths.

Sheet shift

\Delta y = \frac{(\mu - 1)tD}{d} = \frac{(\mu - 1)t}{\lambda}\,\beta

Worked example

A mica sheet (μ = 1.6) 5 μm thick covers one slit. λ = 600 nm, d = 0.5 mm and D = 1 m. By how many fringes, and how far, does the central maximum move?

Practice this conceptself-check · 4 quick reps

The same idea in a real exam question:

JEE Mains · 2026 · 21 Jan 2026 Shift 1 · Q11Moderate

Example 3 · Wave Optics · Double Slit Intensity and Slab Shifts

In a double slit experiment the distance between the slits is 0.1 cm and the screen is placed at 50 cm from the slits plane. When one slit is covered with a transparent sheet having thickness

t

and refractive index

n( = 1.5)

, the central fringe shifts by 0.2 cm. The value of t is

\_\_\_\_

cm.

Towards the covered slit

The sheet lengthens that path, so the equal-path point moves to the side where the geometric path is shorter: the covered side.

Use μ − 1, not μ

The sheet replaces the same thickness of air, so it adds (μ − 1)t. Using μt gives a shift about three times too large for glass.

The fringes do not narrow

The sheet only moves the pattern. β = λD/d is unchanged, so an option that changes the fringe width is wrong.

Drill 6 more on a thin sheet over one slit

Summary — formulas & gotchas at a glance

A revision cheat-sheet for the formulas and gotchas above. Click any concept name to jump back to its full explanation.

Formulas (3)

Path difference from the geometry
Path difference
$\Delta x = \frac{yd}{D}, \qquad \Delta x_{\text{opposite a slit}} = \frac{d^{2}}{2D}, \qquad \sqrt{D^{2} + d^{2}} - D \approx \frac{d^{2}}{2D}$
Intensity from path and phase
Double-slit intensity
$I = 4I_{0}\cos^{2}\frac{\phi}{2} = I_{\max}\cos^{2}\frac{\pi\,\Delta x}{\lambda}, \qquad \phi = \frac{2\pi}{\lambda}\Delta x$
A thin sheet over one slit
Sheet shift
$\Delta y = \frac{(\mu - 1)tD}{d} = \frac{(\mu - 1)t}{\lambda}\,\beta$

Watch out for (9)

Opposite a slit is y = d/2
→ Path difference from the geometry
yd/D needs a far screen
→ Path difference from the geometry
Know which I₀ the question means
→ Path difference from the geometry
Half the phase inside the cosine
→ Intensity from path and phase
I₀ for one slit means I_max = 4I₀
→ Intensity from path and phase
Path is not phase
→ Intensity from path and phase
Towards the covered slit
→ A thin sheet over one slit
Use μ − 1, not μ
→ A thin sheet over one slit
The fringes do not narrow
→ A thin sheet over one slit

Test yourself on Wave Optics

20 past JEE Mains questions from this chapter, timed at 48 minutes and marked the way the exam marks it. You see your score and every answer the moment you finish. Free to start.

Take the Wave Optics chapter test Drill all 20 questions