JEE Mains Physics · Wave Optics

Single-Slit Diffraction and Resolving Power

A single slit of width a gives dark fringes where a sin θ = nλ, a central bright band twice as wide as the rest, and a round aperture limits how finely a telescope or microscope can separate two points.

All Wave Optics notes JEE Mains Physics strategy

Why this matters

Eighteen PYQs, eight of them multiple choice, and four from 2026; eight came in 2024 alone. Eight locate the dark fringes: the distance between two of them, or the slit width that puts one at a given angle. Six ask for the width of the central maximum, two of them by counting the double-slit fringes that fit inside it. Four are about resolving power: a telescope, a microscope in oil and a pinhole.

Concept 1 of 3: Dark fringes of a single slit

Split the slit into two halves. If the edge-to-edge path difference a sin θ is one wavelength, each point in the top half has a partner in the bottom half exactly half a wave behind, and the pairs cancel. So a sin θ = λ is dark, not bright. That is the reverse of the double slit, where a path of λ gives a bright fringe.

Definition

Minima: $a\sin\theta = n\lambda$ , n = 1, 2, 3, … (n = 0 is the bright centre).
On a screen at distance D: $y_{n} = \dfrac{n\lambda D}{a}$ . First to third minimum (same side): $\dfrac{2\lambda D}{a}$ .
Secondary maxima lie roughly halfway: $a\sin\theta \approx \left(n + \tfrac{1}{2}\right)\lambda$ , the first at $\dfrac{3\lambda}{2}$ .
For large angles keep $\sin\theta$ ; the small-angle form $y = n\lambda D/a$ is only for small $\theta$ .
"Angular divergence" of a pair of minima usually means the full angle between them, $2\theta$ ; say which reading you use.

Single-slit minima

a\sin\theta = n\lambda, \qquad y_{n} = \frac{n\lambda D}{a}, \qquad y_{3} - y_{1} = \frac{2\lambda D}{a}

Worked example

Light of 500 nm falls on a slit 0.25 mm wide, with the screen 1.5 m away. Find the distances of the first and second minima from the centre, and of the first secondary maximum.

Practice this conceptself-check · 4 quick reps

The same idea in a real exam question:

JEE Mains · 2024 · 29 January 2024 · Q117Moderate

Example 1 · Wave Optics · Single-Slit Diffraction and Resolving Power

In a single slit diffraction pattern, a light of wavelength

6000\text{Å}

is used. The distance between the first and third minima in the diffraction pattern is found to be

3\text{ }mm

when the screen is placed 50

cm

away from slits. The width of the slit is _____

\times10^{- 4}\text{ }m

nλ is dark for a single slit

In the double slit, a path of nλ is bright. For one slit, a sin θ = nλ is a minimum. Mixing the two puts every fringe in the wrong place.

First to third is two steps

The first and third minima are 2λD/a apart, not 3λD/a. Subtract positions, do not read off the higher order.

Angle of one minimum or the angle between two?

"Angular divergence" can mean θ or 2θ. Check the options: they often contain both.

Drill 7 more on dark fringes of a single slit

Concept 2 of 3: Width of the central maximum

The central bright band runs from the first dark fringe on one side to the first on the other, so it is twice as wide as the gap between later dark fringes. A narrower slit spreads the light more. In a double slit, each slit's own diffraction sets an envelope, and the interference fringes are seen inside it.

Definition

Angular width of the central maximum: $\dfrac{2\lambda}{a}$ (small angles).
Linear width on a screen at D: $\dfrac{2\lambda D}{a}$ ; in the focal plane of a lens of focal length f: $\dfrac{2\lambda f}{a}$ .
Each secondary maximum is half as wide: $\dfrac{\lambda D}{a}$ .
For large angles use $\sin\theta = \lambda/a$ and double the angle for the full spread.
Double slit with slits of width a and gap d: the central maximum, $2\lambda D/a$ , holds $\dfrac{2\lambda D/a}{\lambda D/d} = \dfrac{2d}{a}$ fringe widths. JEE counts this as the number of bright fringes inside it.

Central maximum

\theta_{\text{width}} = \frac{2\lambda}{a}, \qquad W = \frac{2\lambda D}{a}\ \left(\text{or } \frac{2\lambda f}{a}\right), \qquad N = \frac{2d}{a}

Worked example

Light of 450 nm falls on a slit 0.3 mm wide, and the pattern is formed in the focal plane of a lens of focal length 40 cm. Find the angular and linear widths of the central maximum.

Practice this conceptself-check · 4 quick reps

The same idea in a real exam question:

JEE Mains · 2024 · 1 February 2024 · Q20Moderate

Example 2 · Wave Optics · Single-Slit Diffraction and Resolving Power

A monochromatic light of wavelength

6000\text{Å}

is incident on the single slit of width

0.01\text{ }mm

. If the diffraction pattern is formed at the focus of the convex lens of focal length

20\text{ }cm

, the linear width of the central maximum is:

Twice λD/a, not λD/a

The central maximum spans from the first minimum on one side to the first on the other: 2λD/a. The single-step λD/a is the width of a secondary maximum.

a, not d

The single-slit width a sets the envelope; the gap d sets the double-slit fringes. Using λD/d for a central maximum mixes the two patterns.

Large angles need the sine

When λ/a is not small, as with microwaves on a slit a few wavelengths wide, find θ from sin θ = λ/a and double it.

Drill 5 more on width of the central maximum

Concept 3 of 3: Resolving power of a telescope and a microscope

A round lens is a round hole, so it spreads each point of light into a small disc. Two stars can be told apart only if their discs do not overlap too much. A wider lens makes smaller discs, and shorter light does the same. A microscope gains from oil under the lens, which shortens the wavelength in the gap.

Definition

Circular aperture of diameter D: first dark ring at $\sin\theta = \dfrac{1.22\lambda}{D}$ .
Telescope: limit of resolution $\Delta\theta = \dfrac{1.22\lambda}{D}$ ; resolving power is $\dfrac{1}{\Delta\theta}$ . A bigger objective resolves finer detail.
Microscope: resolving power $\dfrac{2\mu\sin\theta}{1.22\lambda}$ . Oil of index μ multiplies it by μ.
Smallest separation resolved at distance R: $R\,\Delta\theta$ .
A larger pinhole gives a smaller, brighter diffraction pattern.

Resolution

\Delta\theta = \frac{1.22\lambda}{D}, \qquad \text{RP}_{\text{microscope}} = \frac{2\mu\sin\theta}{1.22\lambda}

Worked example

A telescope has an objective 1.5 m across and is used with light of 600 nm. Find its limit of resolution, and the smallest separation it can resolve on the Moon,

3.8 \times 10^{8}

m away.

Practice this conceptself-check · 4 quick reps

The same idea in a real exam question:

JEE Mains · 2026 · 4 Apr 2026 Shift 1 · Q19Moderate

Example 3 · Wave Optics · Single-Slit Diffraction and Resolving Power

A telescope with objective diameter R is used to observe a distant star emitting light of wavelength 500 nm , at a resolution of

5 \times10^{- 7}

radian. The value of

R

\_\_\_\_

cm .

Limit and power are opposites

The limit of resolution is an angle, 1.22λ/D; smaller is better. The resolving power is its reciprocal; larger is better.

Keep the 1.22

A round aperture gives 1.22λ/D, not λ/D. Dropping the factor lands on a distractor about 20% off.

Oil helps a microscope, not a telescope

The microscope's resolving power has μ in it, so oil raises it. A telescope's limit depends only on λ and the objective diameter.

Drill 3 more on resolving power of a telescope and a microscope

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)

Dark fringes of a single slit
Single-slit minima
$a\sin\theta = n\lambda, \qquad y_{n} = \frac{n\lambda D}{a}, \qquad y_{3} - y_{1} = \frac{2\lambda D}{a}$
Width of the central maximum
Central maximum
$\theta_{\text{width}} = \frac{2\lambda}{a}, \qquad W = \frac{2\lambda D}{a}\ \left(\text{or } \frac{2\lambda f}{a}\right), \qquad N = \frac{2d}{a}$
Resolving power of a telescope and a microscope
Resolution
$\Delta\theta = \frac{1.22\lambda}{D}, \qquad \text{RP}_{\text{microscope}} = \frac{2\mu\sin\theta}{1.22\lambda}$

Watch out for (9)

nλ is dark for a single slit
→ Dark fringes of a single slit
First to third is two steps
→ Dark fringes of a single slit
Angle of one minimum or the angle between two?
→ Dark fringes of a single slit
Twice λD/a, not λD/a
→ Width of the central maximum
a, not d
→ Width of the central maximum
Large angles need the sine
→ Width of the central maximum
Limit and power are opposites
→ Resolving power of a telescope and a microscope
Keep the 1.22
→ Resolving power of a telescope and a microscope
Oil helps a microscope, not a telescope
→ Resolving power of a telescope and a microscope

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