Diffraction

Master single slit diffraction, diffraction grating, and resolving power for JEE Physics

12 min read

Prerequisites

Before studying this topic, make sure you understand:

The Hook: Why Shadows Aren’t Perfect

Connect: Real Life → Physics

Why can you hear someone around a corner but not see them? Why do you see colorful patterns when looking at a distant street light through a thin cloth? Why can’t we see atoms even with the most powerful optical microscope?

In Interstellar, Cooper says “We need to think beyond three dimensions.” Similarly, light doesn’t just travel in straight lines (ray optics) — it bends around corners and creates patterns when passing through small openings. This is diffraction, and it sets the ultimate limit on what we can see!

Even Spider-Man’s spider-sense can’t see around corners, but light can (to some extent)!

JEE Weightage: MEDIUM-HIGH — Expect 1-2 questions in JEE Main, 1 in Advanced (especially resolving power)

The Core Concept

Diffraction is the bending of waves around obstacles or through openings when the size of the obstacle/opening is comparable to the wavelength.

Why Does Diffraction Occur?

According to Huygens’ Principle, every point on a wavefront acts as a source of secondary wavelets. When a wave passes through a slit:

  • Different parts of the slit act as coherent sources
  • These sources interfere with each other
  • Result: Light “spreads out” and creates a pattern of bright and dark regions

In simple terms: Light doesn’t just go straight through — it spreads out and interferes with itself!

Difference: Interference vs Diffraction

PropertyInterferenceDiffraction
Number of sourcesTwo or more separate sourcesSingle wavefront (many points on it)
Fringe widthUsually equalCentral bright is wider than others
IntensityAll maxima roughly equalIntensity decreases for higher order
ExampleYDSE (two slits)Single slit pattern

Reality: Diffraction and interference often occur together! YDSE pattern is actually interference + diffraction combined.

Single Slit Diffraction

Experimental Setup

Light of wavelength $\lambda$ passes through a single slit of width $a$. A screen is placed at distance $D$.

Pattern observed:

  • Wide central bright fringe (maximum intensity at center)
  • Alternating dark and bright fringes on both sides
  • Intensity decreases rapidly for higher order fringes

Condition for Minima (Dark Fringes)

$$\boxed{a\sin\theta = n\lambda}$$

Where:

  • $a$ = slit width
  • $\theta$ = angular position of dark fringe
  • $n = 1, 2, 3, ...$ (order of minimum)

Position of minima on screen:

$$y_n = \frac{n\lambda D}{a}$$

For small angles: $\sin\theta \approx \tan\theta = \frac{y}{D}$

Width of Central Maximum

The central bright fringe extends from first minimum on one side to first minimum on other side:

$$\boxed{\text{Width of central maximum} = 2y_1 = \frac{2\lambda D}{a}}$$

Compare with YDSE: In YDSE, fringe width $\beta = \frac{\lambda D}{d}$. In single slit, central maximum is twice as wide, and $a$ is in denominator (not $d$).

Condition for Maxima (Bright Fringes)

Secondary maxima occur approximately at:

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

Where $n = 1, 2, 3, ...$

Note: This is approximate. Exact positions require solving a transcendental equation.

Intensity Distribution

$$\boxed{I = I_0 \left(\frac{\sin\beta}{\beta}\right)^2}$$

Where $\beta = \frac{\pi a \sin\theta}{\lambda}$

  • At $\beta = 0$ (center): $I = I_0$ (maximum)
  • At $\beta = \pi, 2\pi, 3\pi, ...$: $I = 0$ (minima)
Central Maximum is Brightest

Unlike YDSE (where all bright fringes have equal intensity), in single slit:

  • Central maximum is brightest
  • Secondary maxima get progressively dimmer
  • Intensity falls off as $\frac{1}{n^2}$ approximately

Diffraction Grating

A diffraction grating is a device with a large number of parallel, equally spaced slits (typically thousands per cm).

Grating Equation

For a grating with slit separation $d$, bright fringes (principal maxima) occur at:

$$\boxed{d\sin\theta = n\lambda}$$

Where:

  • $d$ = grating element (distance between adjacent slits)
  • $\theta$ = angle of diffraction
  • $n = 0, 1, 2, 3, ...$ (order of spectrum)

Memory Trick: “Don’t Sin Normally with Light” → $d\sin\theta = n\lambda$

Number of Lines per Unit Length

If there are $N$ lines per meter:

$$d = \frac{1}{N}$$

Example: 5000 lines/cm = 500,000 lines/m → $d = \frac{1}{500000} = 2 \times 10^{-6}$ m

Maximum Order of Spectrum

Since $\sin\theta \leq 1$:

$$n_{max} = \left\lfloor \frac{d}{\lambda} \right\rfloor$$

(Greatest integer less than or equal to $\frac{d}{\lambda}$)

In simple terms: How many complete spectra can we see? Depends on grating spacing and wavelength.

Absent Spectra

If both single slit diffraction and grating interference are considered, some orders may be missing (absent).

Condition for absent order: If $\frac{a}{d} = \frac{p}{q}$ (in lowest terms), then orders $n = q, 2q, 3q, ...$ are absent.

Example: If $a = \frac{d}{2}$, then $n = 2, 4, 6, ...$ (even orders) are missing!

Resolving Power

The ability to distinguish between two closely spaced objects or wavelengths.

Rayleigh’s Criterion

Two point sources are just resolved when the central maximum of one coincides with the first minimum of the other.

Angular separation for resolution:

$$\boxed{\theta_{min} = \frac{1.22\lambda}{D}}$$

Where:

  • $D$ = diameter of circular aperture (like telescope lens or eye pupil)
  • Factor 1.22 comes from circular aperture (for slit, it’s just $\frac{\lambda}{a}$)

Resolving Power of Telescope

$$\boxed{R_T = \frac{1}{\theta_{min}} = \frac{D}{1.22\lambda}}$$

In simple terms: Bigger telescope (larger $D$) → better resolving power → can see finer details!

This is why:

  • Hubble telescope has a 2.4 m mirror (high resolving power)
  • Radio telescopes need to be HUGE (radio waves have large $\lambda$)

Resolving Power of Microscope

$$\boxed{R_M = \frac{2\mu\sin\theta}{1.22\lambda}}$$

Where:

  • $\mu$ = refractive index of medium between object and lens
  • $\theta$ = half-angle of cone of light entering objective

This is why:

  • Electron microscopes exist (electrons have tiny wavelength)
  • We can’t see atoms with optical microscope (visible light $\lambda$ too large)

Resolving Power of Grating

The ability to distinguish between two nearby wavelengths:

$$\boxed{R_G = nN}$$

Where:

  • $n$ = order of spectrum
  • $N$ = total number of lines in the grating

In simple terms: More lines + higher order = better wavelength resolution!

Interstellar Telescope
In Interstellar, they use powerful telescopes to see distant phenomena. The resolving power determines how far apart two stars must be to see them as separate — not just one blob of light!

Comparison: Single Slit vs Double Slit vs Grating

FeatureSingle SlitDouble Slit (YDSE)Diffraction Grating
Number of slits12Many (thousands)
PatternWide central, dim sidesEqually spaced, equal intensitySharp, bright lines
Formula for maxima$(n+\frac{1}{2})\lambda$ (approx)$n\lambda$$n\lambda$
Formula for minima$n\lambda$$(2n-1)\frac{\lambda}{2}$Many minima between maxima
Central maximum width$\frac{2\lambda D}{a}$——
Fringe widthDecreasingConstant—
Intensity distributionDecreasingConstantDecreasing
SharpnessBroad fringesModerateVery sharp

Memory Tricks & Patterns

Mnemonic for Grating Equation

Memory Trick: “Drama Starts with New Lines” → $d\sin\theta = n\lambda$

Mnemonic for Resolving Power

Memory Trick: “Resolve with nice Numbers” → $R = nN$ (for grating)

Pattern Recognition

Diffraction Patterns
  1. Slit width increases → Central maximum gets narrower
  2. Wavelength increases → Pattern spreads out (wider)
  3. More slits in grating → Lines become sharper
  4. Higher order spectrum → Spreads out more (larger $\theta$)
  5. Circular aperture → Factor 1.22 appears (not for rectangular slit)

Quick Decision Tree

Diffraction problem?
├─ Single slit? → Minima at a·sinθ = nλ
├─ Grating? → Maxima at d·sinθ = nλ
├─ Resolving power?
│   ├─ Telescope/Microscope? → R = D/(1.22λ)
│   └─ Grating? → R = nN
└─ Width of central max? → 2λD/a

Common Mistakes to Avoid

Trap #1: Single Slit vs YDSE Formula

Wrong: Using $d\sin\theta = n\lambda$ for single slit minima

Correct:

  • Single slit minima: $a\sin\theta = n\lambda$ (uses slit width $a$)
  • YDSE/Grating maxima: $d\sin\theta = n\lambda$ (uses slit separation $d$)

Opposite roles! For single slit, this gives minima; for grating, it gives maxima.

Trap #2: Central Maximum Width

Wrong: Central maximum width = $\frac{\lambda D}{a}$ (like YDSE fringe width)

Correct: Central maximum width = $\frac{2\lambda D}{a}$ (factor of 2!)

It extends from first minimum to first minimum on opposite side.

Trap #3: Grating Lines Confusion

Wrong: Using number of lines directly as $d$

Correct: If $N$ lines per cm, then $d = \frac{1}{N}$ (in cm)

Example: 5000 lines/cm → $d = \frac{1}{5000}$ cm = $2 \times 10^{-4}$ cm

Convert to meters!

Trap #4: Resolving Power Factor

Wrong: Using $\theta_{min} = \frac{\lambda}{D}$ for circular aperture

Correct: $\theta_{min} = \frac{1.22\lambda}{D}$ (factor 1.22 for circular!)

For rectangular slit, it’s $\frac{\lambda}{a}$ (no 1.22)

Practice Problems

Level 1: Foundation (NCERT-type)

Problem 1: Single Slit First Minimum

Light of wavelength 600 nm falls on a slit of width 0.3 mm. A screen is placed 1 m away. Find the position of the first dark fringe.

Solution: For first minimum: $n = 1$

$$y_1 = \frac{n\lambda D}{a} = \frac{1 \times 6 \times 10^{-7} \times 1}{3 \times 10^{-4}} = 2 \times 10^{-3} \text{ m}$$

Answer: 2 mm from center

Problem 2: Width of Central Maximum

For the same setup as Problem 1, find the width of the central maximum.

Solution:

$$\text{Width} = 2y_1 = 2 \times 2 = 4 \text{ mm}$$

Answer: 4 mm

Problem 3: Grating Equation

A diffraction grating has 5000 lines/cm. At what angle will the first order maximum occur for light of wavelength 500 nm?

Solution: $N = 5000$ lines/cm $= 500,000$ lines/m

$$d = \frac{1}{N} = \frac{1}{500,000} = 2 \times 10^{-6} \text{ m}$$

For first order: $n = 1$

$$d\sin\theta = n\lambda$$ $$\sin\theta = \frac{\lambda}{d} = \frac{5 \times 10^{-7}}{2 \times 10^{-6}} = 0.25$$ $$\theta = \sin^{-1}(0.25) \approx 14.5°$$

Answer: 14.5°

Level 2: JEE Main

Problem 4: Maximum Order

A grating with 4000 lines/cm is illuminated with light of wavelength 625 nm. What is the highest order spectrum that can be observed?

Solution:

$$d = \frac{1}{4000 \times 100} = 2.5 \times 10^{-6} \text{ m}$$

Maximum order when $\sin\theta = 1$:

$$n_{max} = \frac{d}{\lambda} = \frac{2.5 \times 10^{-6}}{625 \times 10^{-9}} = \frac{2.5}{0.625} = 4$$

Answer: 4th order (i.e., $n = 0, 1, 2, 3, 4$ → five spectra visible)

Problem 5: Slit Width from Central Maximum

In a single slit diffraction experiment, the width of the central maximum is 4 cm when screen is 2 m away and wavelength is 600 nm. Find the slit width.

Solution:

$$\text{Width} = \frac{2\lambda D}{a}$$ $$4 \times 10^{-2} = \frac{2 \times 6 \times 10^{-7} \times 2}{a}$$ $$a = \frac{2 \times 6 \times 10^{-7} \times 2}{4 \times 10^{-2}} = \frac{24 \times 10^{-7}}{4 \times 10^{-2}} = 6 \times 10^{-5} \text{ m}$$

Answer: 0.06 mm or 60 μm

Problem 6: Resolving Power of Telescope

A telescope has an objective lens of diameter 1 m. What is its resolving power for light of wavelength 500 nm?

Solution:

$$R = \frac{D}{1.22\lambda} = \frac{1}{1.22 \times 5 \times 10^{-7}} = \frac{10^7}{6.1} \approx 1.64 \times 10^6$$

Answer: $1.64 \times 10^6$ (can resolve objects 1.64 million times closer than their distance)

Level 3: JEE Advanced

Problem 7: Absent Spectra

A diffraction grating has slit width $a$ equal to twice the separation $d$ between adjacent slits (i.e., $a = 2d$). Which orders of spectra will be absent?

Solution: For absent order: diffraction minimum coincides with grating maximum

Single slit minima: $a\sin\theta = m\lambda$ Grating maxima: $d\sin\theta = n\lambda$

For coincidence: $\frac{a}{d} = \frac{n}{m}$

Given $a = 2d$, so $\frac{a}{d} = 2 = \frac{n}{m}$

This means $n = 2m$, i.e., $n = 2, 4, 6, 8, ...$ (when $m = 1, 2, 3, ...$)

Answer: 2nd, 4th, 6th, … (all even orders) are absent

Problem 8: Resolving Power of Grating

A grating has 8000 lines and is used in 3rd order. What is its resolving power? Can it resolve wavelengths 500.0 nm and 500.5 nm?

Solution:

$$R = nN = 3 \times 8000 = 24,000$$

For resolving $\lambda$ and $\lambda + \Delta\lambda$:

$$R = \frac{\lambda}{\Delta\lambda}$$

Required resolving power:

$$R_{req} = \frac{500}{0.5} = 1000$$

Since $24,000 > 1000$, YES, it can easily resolve these wavelengths!

Answer: Resolving power = 24,000; Yes, can resolve

Problem 9: Angular Spread

White light (400 nm to 700 nm) falls normally on a grating with 5000 lines/cm. Find the angular separation between red (700 nm) and violet (400 nm) in the first order spectrum.

Solution:

$$d = \frac{1}{500,000} = 2 \times 10^{-6} \text{ m}$$

For violet (400 nm):

$$\sin\theta_V = \frac{n\lambda_V}{d} = \frac{1 \times 400 \times 10^{-9}}{2 \times 10^{-6}} = 0.2$$ $$\theta_V = \sin^{-1}(0.2) = 11.54°$$

For red (700 nm):

$$\sin\theta_R = \frac{700 \times 10^{-9}}{2 \times 10^{-6}} = 0.35$$ $$\theta_R = \sin^{-1}(0.35) = 20.49°$$

Angular spread:

$$\Delta\theta = \theta_R - \theta_V = 20.49° - 11.54° = 8.95°$$

Answer: ~9° (angular width of first order spectrum)

Quick Revision Box

SituationFormula/Approach
Single slit minima$a\sin\theta = n\lambda$, $n = 1,2,3,...$
Width of central maximum$\frac{2\lambda D}{a}$
Grating maxima$d\sin\theta = n\lambda$, $n = 0,1,2,...$
Grating element from lines$d = \frac{1}{N}$ (N = lines per unit length)
Maximum order$n_{max} = \lfloor\frac{d}{\lambda}\rfloor$
Resolving power (telescope)$R = \frac{D}{1.22\lambda}$
Resolving power (grating)$R = nN$
Rayleigh criterion$\theta_{min} = \frac{1.22\lambda}{D}$ (circular aperture)
Absent ordersIf $\frac{a}{d} = \frac{p}{q}$, then $n = q, 2q, 3q,...$ absent

JEE Exam Strategy

Time-Saving Tips
  1. Identify the setup: Single slit? Grating? Resolving power?
  2. Watch for “lines per cm” → Convert to $d = \frac{1}{N}$ immediately
  3. Factor 1.22 appears only for circular apertures (telescope, microscope pupil)
  4. Maximum order: Quick check: $n_{max} = \frac{d}{\lambda}$ (round down)
  5. Absent spectra: Rarely asked, but easy marks if you know the condition

Common JEE Patterns:

  • Single slit central maximum width (direct formula)
  • Grating equation for finding angle or order
  • Maximum order calculation
  • Resolving power of telescope (factor 1.22 trap!)
  • Grating resolving power (simple: $nN$)

High-Yield Concepts:

  1. Grating equation: $d\sin\theta = n\lambda$ (appears in 90% of grating problems)
  2. Central maximum width: $\frac{2\lambda D}{a}$ (factor of 2!)
  3. Resolving power formulas (know all three: telescope, microscope, grating)

Within Optics

Connected Chapters

Real-World Applications

  • CDs and DVDs: Diffraction grating creates rainbow effect
  • X-ray crystallography: Determining atomic structures
  • Spectroscopy: Identifying elements by spectral lines
  • Astronomy: Resolving power limits what telescopes can see
  • Holography: Recording interference and diffraction patterns

Math Connections

Teacher’s Summary

Key Takeaways
  1. Diffraction is bending of light around obstacles/through openings when size ~ wavelength
  2. Single slit: Minima at $a\sin\theta = n\lambda$ → Central max is widest
  3. Grating: Maxima at $d\sin\theta = n\lambda$ → Sharp, bright spectral lines
  4. Resolving power sets limits on what we can distinguish:
    • Telescope/Microscope: $R = \frac{D}{1.22\lambda}$ (bigger diameter = better)
    • Grating: $R = nN$ (more lines, higher order = better)
  5. Central maximum width = $\frac{2\lambda D}{a}$ (factor of 2, don’t forget!)
  6. Maximum order from grating: $n_{max} = \frac{d}{\lambda}$ (limited by $\sin\theta \leq 1$)

“Diffraction shows light is a wave. Single slit spreads it out, grating sharpens it up, and resolving power tells us the limits of what we can see!”

JEE Success Mantra: Grating problems are straightforward — memorize $d\sin\theta = n\lambda$ and $R = nN$. Single slit: remember the factor of 2 in central maximum width. Practice 15 problems and these become automatic!