MHT-CET Physics · Wave Optics

Single-Slit Diffraction and Resolving Power

Light through a single slit of width a spreads into a bright central band bounded by minima at a sin θ = ±λ, with fainter secondary maxima near a sin θ = (n + ½)λ; the same spreading limits how finely a microscope or an eye can resolve.

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Why this matters

34 PYQs, seven HARD. Three shapes: the minima and the width of the central maximum (and what happens when the slit or the wavelength changes), the secondary maxima — where they fall and when two wavelengths' maxima coincide — and the limit of resolution.

Concept 1 of 3: Minima and the Central Maximum

Pair each point in the top half of the slit with one in the bottom half: when their path difference is λ/2 they cancel, and so does the whole slit. That happens at a sin θ = λ, the first minimum. A narrower slit or longer wavelength pushes it out, widening the central band.

Definition

Minima: $a\sin\theta = n\lambda$ ; on a screen at distance D, $y_n = \dfrac{n\lambda D}{a}$ .
Central maximum: linear width $\dfrac{2\lambda D}{a}$ , angular width $\dfrac{2\lambda}{a}$ , twice as wide as the other bands.
Slit width halved and $\lambda \times 1.5$ : width $\times 3$ . Width $\propto \lambda$ : 30% less angular width means 30% shorter $\lambda$ .
Doubling the slit: central intensity $\times 4$ , angular width $\times \dfrac{1}{2}$ .
The bands have unequal widths and unequal intensities.

Single-slit minima

a\sin\theta = n\lambda, \qquad W_{\text{central}} = \frac{2\lambda D}{a}

Worked example

\lambda = 500

nm, slit width 0.25 mm, screen 1.5 m away. Width of the central maximum?

Practice this conceptself-check · 3 quick reps

The same idea in a real exam question:

MHT-CET · 2024 · 11th May Shift 2 · Q47Easy

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

A beam of light of wavelength 600 nm from a distant source falls on a single slit 1 mm wide and the resulting diffraction pattern is observed on a screen 2 m away. The distance between the first dark fringe on either side of the central bright fringe is

Half-width or full width

\frac{\lambda D}{a}

is the distance from the centre to the first minimum. 'Between the first minima on either side' is twice that,

\frac{2\lambda D}{a}

Drill 19 more on minima and the central maximum

Concept 2 of 3: Secondary Maxima

Between the minima the pattern rises to fainter maxima, roughly halfway along: the nth secondary maximum sits near a sin θ = (n + ½)λ. Two wavelengths' maxima coincide where those positions match.

Definition

$n$ th secondary maximum: $a\sin\theta = \left(n + \dfrac{1}{2}\right)\lambda$ , $y = \dfrac{(2n + 1)\lambda D}{2a}$ .
First minimum at $30^\circ$ ( $a = 2\lambda$ ): first secondary maximum at $\sin\theta = \dfrac{3}{4}$ .
Coincidence: $n$ th maximum of $\lambda_1$ with $m$ th of $\lambda_2$ : $(2n + 1)\lambda_1 = (2m + 1)\lambda_2$ .
Two close wavelengths: $\Delta y = \dfrac{3D}{2a}\Delta\lambda$ for the first secondary maxima.

Secondary maxima

a\sin\theta = \left(n + \tfrac{1}{2}\right)\lambda

Worked example

The 3rd secondary maximum of 500 nm light coincides with the 2nd secondary maximum of light of wavelength λ. Find λ.

Practice this conceptself-check · 1 quick reps

The same idea in a real exam question:

MHT-CET · 2025 · 25 April Shift I · Q23Moderate

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

A single slit diffraction pattern is formed with light of wavelength

6384\text{Å}

. The second secondary maximum for this wavelength coincides with the third secondary maximum in the pattern for light of wavelength '

\lambda_{0}

'. The value of '

\lambda_{0}

' is

Using nλ for a maximum

In SINGLE-slit diffraction

a\sin\theta = n\lambda

gives the MINIMA — the opposite of the double-slit rule. Secondary maxima need the extra half.

Drill 9 more on secondary maxima

Concept 3 of 3: Resolving Power

Every point of an object is imaged as a small diffraction disc; two points blur into one when their discs overlap too much. Shorter light makes smaller discs, and a denser medium between object and lens shortens the light — so both sharpen the image.

Definition

Limit of resolution $\propto \lambda$ ; resolving power $= \dfrac{2\mu\sin\theta}{1.22\lambda}$ for a microscope.
Improve it: shorter wavelength, larger aperture, or a higher-index medium (oil immersion). A longer wavelength or smaller objective makes it worse.
The eye: $\Delta\theta = \dfrac{1.22\lambda}{D_{\text{pupil}}}$ ; at viewing distance $L$ the smallest separation is $L\,\Delta\theta$ .

Angular resolution

\Delta\theta = \frac{1.22\lambda}{D}

Worked example

A microscope just resolves 0.2 μm with 600 nm light. With 450 nm light?

Practice this conceptself-check · 1 quick reps

The same idea in a real exam question:

MHT-CET · 2023 · 2nd May Shift 2 · Q36Easy

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

Two points separated by a distance of 0.1 mm can just be seen in a microscope when light of wavelength 6000 Å is used. If the light of wavelength 4800 Å is used, the limit of resolution will become

A longer wavelength sees finer detail

Resolving power goes as

\frac{1}{\lambda}

: longer waves blur more. 'Increase the wavelength' is the planted wrong way to improve a microscope.

Drill 3 more on resolving power

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)

Minima and the Central Maximum
Single-slit minima
$a\sin\theta = n\lambda, \qquad W_{\text{central}} = \frac{2\lambda D}{a}$
Secondary Maxima
Secondary maxima
$a\sin\theta = \left(n + \tfrac{1}{2}\right)\lambda$
Resolving Power
Angular resolution
$\Delta\theta = \frac{1.22\lambda}{D}$

Watch out for (3)

Half-width or full width
→ Minima and the Central Maximum
Using nλ for a maximum
→ Secondary Maxima
A longer wavelength sees finer detail
→ Resolving Power

Test yourself on Wave Optics

20 past MHT-CET questions from this chapter, timed at 18 minutes and marked the way the exam marks it. You see your score and every answer the moment you finish. Free to start.

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Related notes

Young's Double Slit — fringe positions →