Double-slit diffraction is the combined optical phenomenon of interference arising from two coherent sources and diffraction occurring due to the finite width of each individual slit. The resulting intensity distribution is a product of single-slit diffraction envelope and double-slit interference fringes.
UP TGT ScienceUP PGT PhysicsStudy Notes + Practice
Scope and prerequisites
TGT core
Qualitative understanding of intensity distribution, conditions for missing orders, positions of principal maxima and minima, and comparison with pure double-slit interference.
PGT extension
Rigorous mathematical derivation of the resultant intensity pattern combining diffraction and interference factors, analytical determination of missing (absent) spectra, and angular width calculations.
Prerequisites
Huygens-Fresnel Principle
Fraunhofer Diffraction at a Single Slit
Superposition Principle of Waves
Superposition of Interference and Diffraction
In a double-slit arrangement where each slit has a finite width a and are separated by center-to-center distance d, both diffraction and interference occur simultaneously.
The overall intensity pattern is formed by modulating the rapid interference fringes by the slow-varying single-slit diffraction envelope.
The diffraction factor governs the overall amplitude envelope, while the interference factor determines the rapid oscillations within that envelope.
Resultant Amplitude and Intensity
The amplitude due to a single slit of width a at an angle θ is given by $A = A_{0}(\sin \alpha/\alpha)$, where $\alpha = (\pia \sin \theta)/\lambda$.
The resultant amplitude from two identical slits separated by distance d introduces a phase difference 2β, where $\beta = (\pid \sin \theta)/\lambda$.
The resultant intensity is expressed as $I = I_{0} (\sin \alpha / \alpha)^{2} \cos^{2} \beta$.
Conditions for Maxima and Minima
Interference maxima occur when d sin $\theta = n\lambda$, where n = 0, ±1, ±2, ...
Diffraction minima (zeroes of the envelope) occur when a sin $\theta = m\lambda$, where $m = \pm 1$, ±2, ±3, ... (excluding m = 0 which corresponds to the central maximum).
The angular width of the central diffraction maximum is twice that of the single slit: 2$\theta = 2\lambda/a$.
Missing Orders in Double-Slit Diffraction
A missing order (or absent spectrum) occurs when an interference maximum coincides with a diffraction minimum.
The mathematical condition is given by d sin $\theta = n\lambda$ and a sin $\theta = m\lambda$, which simplifies to d/$a = n/m$.
If the ratio d/a is an integer (e.g., d = 2a), then the 2nd, 4th, 6th... interference orders will be completely missing from the pattern.
Derivations
Derivations of Intensity Distribution in Double-Slit Diffraction
Conditions: Fraunhofer diffraction regime using two parallel rectangular slits of width a and center separation d illuminated by monochromatic light of wavelength λ normal to the screen.
Consider two identical slits, each of width a, spanning from $x = -a/2$ to a/2 for the first slit, and $x = d-a/2$ to d+a/2 for the second slit.
Integrate the complex amplitude over the aperture of the first slit to obtain the single-slit amplitude component: E1 = A0 (sin α / α), where $\alpha = (\pia \sin \theta)/\lambda$.
Account for the path difference between corresponding points in the two slits, which gives a phase difference $\phi = 2\beta = 2(\pid \sin \theta)/\lambda between the secondary wavelets originating from the centers of the two slits$.
Sum the complex amplitudes from both slits using phasor addition or direct integration: Enet = E1 + E2 = A0 (sin α / α) [1 + ei 2β].
Multiply the net amplitude by its complex conjugate to find the resultant intensity: $I = I_{0} (\sin \alpha / \alpha)^{2} \cos^{2} \beta$.
$I_0 = maximum intensity at central peak$, alpha and \beta = phase angles W/m^2
Monochromatic light, equal slit widths
Interference Maxima Condition
$$d \sin \theta = n \lambda$$
n = 0, pm 1, pm 2, \dots (order of interference) dimensionless integer
Principal maxima
Diffraction Minima Condition
$$a \sin \theta = m \lambda$$
m = pm 1, pm 2, pm 3, \dots (order of diffraction minimum) dimensionless integer
Zeroes of envelope
Missing Order Condition
$$\frac{d}{a} = \frac{n}{m}$$
n = interference order, m = diffraction minimum order dimensionless
Coincidence of interference maxima and diffraction minima
Angular width of central diffraction maximum
$$2\theta = \frac{2lambda}{a}$$
lambda = wavelength, a = slit width radians
Small angle approximation
Number of interference fringes within central diffraction envelope
$$N = \frac{2d}{a}$$
d = slit separation, a = slit width dimensionless
Approximation for small angles
Graphs and experiments
Intensity vs. Angle (or Position) Graph for Double-Slit Diffraction
Draw: A bell-shaped envelope curve representing (sin alpha / alpha)^2, enclosing rapid oscillatory interference fringes representing cos^2 \beta. Show zero intensity where the envelope touches the axis.
Interpret: Visualizes how the single-slit diffraction envelope modulates the double-slit interference fringes, causing certain interference peaks to vanish.
Exam use: Frequently asked in PGT/TGT to identify missing orders or sketch the combined diffraction-interference pattern.
Young's Double Slit vs. Double-Slit Diffraction Comparison
Draw: Two side-by-side graphs: one with uniform intensity peaks (ideal Young's slits where a -> 0) and one with envelope modulation (finite slit width a).
Interpret: Demonstrates that ideal interference assumes point sources, whereas real optical slits exhibit diffraction attenuation.
Exam use: Conceptual questions distinguishing pure interference from combined interference-diffraction.
Optical Bench Setup for Fraunhofer Diffraction
Draw: A linear layout showing a sodium lamp source, a collimating lens L1, a double-slit aperture, a focusing lens L2, and a viewing screen or eyepiece at the focal plane.
Interpret: Converts diverging wavefronts into plane wavefronts (L1) and focuses parallel diffracted rays onto a screen (L2).
Exam use: Experimental setup questions in practical-oriented theory exams.
Solved numericals
TGT core
In a double-slit diffraction pattern, the slit width is 0.1 mm and the separation between the centers of two slits is 0.5 mm. Find which order of interference maximum will be missing inside the central diffraction maximum.
Identify given parameters: slit width a = 0.1 mm, separation d = 0.5 mm.
For the first missing order inside the central envelope, set m = 1, which gives n = 5.
Answer: 5th interference order
PGT extension
Light of wavelength 500 nm falls normally on a double slit. Each slit has a width of 0.2 mm and they are spaced 1.0 mm apart center-to-center. Calculate the total number of interference fringes contained within the central diffraction envelope.
The central diffraction envelope spans from the first diffraction minimum on the left ($m = -1)$ to the first diffraction minimum on the right (m = 1).
The angle \theta for the first diffraction minimum is given by a sin $\theta = \lambda => \sin \theta = \lambda / a$.
The interference maxima are given by d sin \theta = n lambda. Substitute sin $\theta = \lambda / a$ to find the interference order at the diffraction minimum: d (lambda / a) = n lambd$a => n = d / a$.
This means the 5th interference maximum coincides with the 1st diffraction minimum. Therefore, the interference fringes from $n = -4$ to n = 4 lie entirely inside the central envelope, plus the central peak n = 0.
Total number of complete fringes in the central maximu$m = 2(5 - 1) + 1 = 9$, or count orders n = 0, pm 1, pm 2, pm 3, pm 4 (total 9 fringes).
Answer: 9 interference fringes
PGT extension
An optical arrangement uses slits of width 0.05 mm separated by 0.25 mm. If monochromatic light of wavelength 589 nm is used, determine the ratio of the intensity of the first subsidiary interference maximum to the principal central maximum.
For the central principal maximum, alpha = 0 and \beta = 0, so $I = I_0$.
For the first subsidiary interference maximum, determine its position using \beta approx pi / 2 (or solve d sin $\theta = \lambda / 2$ for first interference peak away from center, though strictly n = 0, 1...). Wait, let's use $\beta = \pi/2$ for the first interference maximum outside n=0 (i.e., n=1, $\beta = \pi d \sin\theta / \lambda = \pi d (\lambda / d)/\lambda = \pi)$. Correction: for n=1, \beta = pi. Let's compute alpha corresponding to n=1.
From d/$a = 0.25 / 0.05 = 5$, we have d = 5a. Thus alph$a = \pi a \sin \theta / \lambda = (\pi a / \lambda) (n \lambda / d) = n \pi (a/d) = n \pi / 5$.
For the first interference maximum (n = 1), alph$a = \pi / 5 radians (or 36 degrees)$, and $\beta = \pi radians (where \cos^2 \beta = 1 if evaluated at peak$, wait, cos^2( pi ) = 1, but for n=1, $\beta = \pi radians gives \cos^2(\pi) = 1)$.
Calculate the diffraction factor: (sin(pi/5) / (pi/5))^2. Since pi/5 rad = 36°, sin(36°) approx 0.5878, pi/5 approx 0.6283.
Multiply by cos^2 \beta (which is 1 at the interference maximum): I / I_0 approx 0.875.
Answer: 0.875
Assumptions and limitations
Assumes Fraunhofer conditions where both the source and the screen are effectively at infinity (achieved using lenses in laboratory setups).
Assumes incident light is perfectly monochromatic and coherent.
Neglects edge effects and finite thickness of the partition between the two slits.
Common misconceptions
Common error: Double-slit diffraction pattern is identical to Young's double-slit interference pattern.
Correct: Young's double-slit pattern assumes infinitely narrow slits (constant intensity envelope), whereas real slits produce diffraction modulation that varies the fringe intensities and causes missing orders.
Finite slit width introduces single-slit diffraction envelope terms (sin alpha / alpha)^2 that modulate the interference fringes.
Common error: Missing orders mean light is absorbed at those positions.
Correct: Light is redistributed, not destroyed; missing orders correspond to points where the interference condition yields a maximum but the diffraction envelope has a zero (minimum).
Destructive interference from individual slit components cancels the light amplitude precisely where an interference maximum is expected.
Quick revision
Total Intensit$y = Single-slit diffraction factor × Double-slit interference factor: I = I_0 (\sin \alpha / \alpha)^2 \cos^2 \beta$.
Missing order condition: d / $a = n / m$.
Central maximum angular width: 2$\theta = 2lambda / a$.
Number of interference fringes in central maximum depends directly on the ratio 2d/a.
Must remember
The overall intensity pattern in double-slit diffraction is given by $I = I_0 \left(\frac{\sin \alpha}{\alpha}\right)^2 \cos^2 \beta$, where $\alpha = \frac{\pi a \sin \theta}{\lambda}$ and $\beta = \frac{\pi d \sin \theta}{\lambda}$.
Interference maxima occur at $d \sin \theta = n \lambda$, while diffraction minima occur at $a \sin \theta = m \lambda$ ($m \neq 0$).
A missing order occurs when an interference maximum coincides with a diffraction minimum, leading to the condition $\frac{d}{a} = \frac{n}{m}$.
The angular width of the central diffraction maximum is given by $2\theta = \frac{2\lambda}{a}$.
Glossary
Slit Width ($a$)The finite width of each individual slit responsible for the single-slit diffraction envelope.
Slit Separation ($d$)The center-to-center distance between the two slits responsible for the rapid interference fringes.
Missing OrderAn absent interference maximum in the double-slit pattern because it coincides with a diffraction minimum.
Single-slit diffraction phase anglealpha = \frac{pi a sin \theta}{lambda} (Fraunhofer approximation)
Interference vs Diffraction in Double-Slit Pattern
Feature
Pure Interference (Young's)
Double-Slit Diffraction
Slit Width
Infinitely narrow ($a \to 0$)
Finite width ($a > 0$)
Envelope Intensity
Constant across all fringes
Modulated by single-slit diffraction factor
Missing Orders
None
Possible when $d/a$ is a rational number
Practice Quiz
Choose an option to reveal the explanation.
1. What is the single-slit diffraction phase angle $\alpha$ defined as in terms of slit width $a$ and diffraction angle $\theta$?
Explanation: The phase angle associated with diffraction from a single slit of width $a$ is $\alpha = \frac{\pi a \sin \theta}{\lambda}$.
2. What is the condition for diffraction minima (zeroes of the diffraction envelope) in a double-slit arrangement?
Explanation: Diffraction minima occur at $a \sin \theta = m\lambda$ where $m = \pm 1, \pm 2, \dots$, excluding $m = 0$ which is the central maximum.
3. The resultant intensity distribution for double-slit diffraction is expressed as:
Explanation: The intensity is a combination of the single-slit diffraction factor $\left(\frac{\sin \alpha}{\alpha}\right)^2$ and the double-slit interference factor $\cos^2 \beta$.
4. What does the double-slit interference phase angle $\beta$ depend upon?
Explanation: The interference phase angle is given by $\beta = \frac{\pi d \sin \theta}{\lambda}$, which depends directly on the center-to-center slit separation $d$.
5. If the ratio of slit separation to slit width is $d/a = 2$, which interference orders will be missing?
Explanation: When $d/a = n/m = 2$, whenever $n$ is a multiple of 2 (i.e., $2, 4, 6, \dots$), the interference maximum coincides with a diffraction minimum, making them missing orders.
6. What is the angular width of the central diffraction maximum in a single slit of width $a$?
Explanation: The central diffraction maximum lies between the first diffraction minima at $\sin \theta = \pm \frac{\lambda}{a}$, giving an angular width of $2\theta = \frac{2\lambda}{a}$.
7. Why are certain orders missing in a double-slit diffraction pattern?
Explanation: Missing orders occur precisely at angles where an interference maximum condition ($d \sin \theta = n\lambda$) meets a diffraction minimum condition ($a \sin \theta = m\lambda$).
8. In the expression $I = I_0 \left(\frac{\sin \alpha}{\alpha}\right)^2 \cos^2 \beta$, what does $\left(\frac{\sin \alpha}{\alpha}\right)^2$ represent?
Explanation: The term $\left(\frac{\sin \alpha}{\alpha}\right)^2$ governs the single-slit diffraction envelope that modulates the intensity of the interference fringes.
9. What happens to the double-slit pattern when the slit width $a$ approaches zero while keeping $d$ constant?
Explanation: As $a \to 0$, the diffraction envelope $\left(\frac{\sin \alpha}{\alpha}\right)^2 \to 1$, reducing the pattern to ideal Young's double-slit interference.
10. If $d = 3a$, which interference orders will be missing in the diffraction pattern?
Explanation: Using $d/a = n/m$, if $d/a = 3$, then for $m = 1, 2, 3$, the missing interference orders are $n = 3m$, which gives $3, 6, 9, \dots$.
11. Principal interference maxima are given by which equation?
Explanation: Principal maxima for interference from two slits separated by distance $d$ occur when $d \sin \theta = n\lambda$, where $n = 0, \pm 1, \pm 2, \dots$
12. What is the approximate number of interference fringes contained within the central diffraction maximum?
Explanation: The number of interference fringes within the central diffraction envelope is approximately given by $N = \frac{2d}{a}$.
13. Which factor primarily determines the rapid oscillations (fringes) inside the diffraction envelope?
Explanation: The rapid oscillations are governed by the $\cos^2 \beta$ term, which originates from the interference between the two slits.
14. Under what approximation are the diffraction and interference equations typically evaluated for Fraunhofer diffraction?
Explanation: Fraunhofer diffraction formulas commonly utilize the small angle approximation for analytical tractability of positions and widths.
15. Can the 0th order (central maximum) of interference ever be a missing order?
Explanation: Diffraction minima are defined for $m = \pm 1, \pm 2, \dots$ (excluding $m = 0$). Thus, the central maximum ($n = 0, m = 0$) is never missing.
PYQ Patterns
These are pattern-based practice questions. Authentic year labels are only used when verified from a source.
1. In a double-slit diffraction experiment, the width of each slit is $a$ and their separation is $d$. If the 4th interference maximum falls precisely on the 1st diffraction minimum, what is the ratio $d/a$?
Explanation: The missing order condition is $\frac{d}{a} = \frac{n}{m}$. Here, $n = 4$ and $m = 1$, so $\frac{d}{a} = \frac{4}{1} = 4$.
2. The intensity distribution in a double-slit diffraction pattern is characterized by:
Explanation: The combined pattern features rapid interference fringes modulated by the slower varying single-slit diffraction envelope.
3. If the slit width $a$ is equal to the slit separation $d$ in a double-slit setup, which interference orders will be missing?
Explanation: If $d = a$, then $\frac{d}{a} = 1 = \frac{n}{m}$. Thus, for every integer $m$, $n = m$, meaning all interference orders corresponding to $n = \pm 1, \pm 2, \dots$ coincide with diffraction minima and are missing.
4. What is the angular half-width of the central diffraction maximum for a slit of width $a$ illuminated by wavelength $\lambda$?
Explanation: The first diffraction minimum occurs at $\sin \theta = \frac{\lambda}{a}$ (or $\theta \approx \frac{\lambda}{a}$), which represents the angular half-width of the central maximum.
5. Which of the following correctly distinguishes Young's double-slit interference from double-slit diffraction?
Explanation: Real slits have finite width, producing single-slit diffraction effects that modulate the interference pattern, unlike idealized Young's slits.
6. In the double-slit intensity formula $I = I_0 \left(\frac{\sin \alpha}{\alpha}\right)^2 \cos^2 \beta$, if $a \to 0$, what does the formula reduce to?
Explanation: As $a \to 0$, $\alpha \to 0$ and $\left(\frac{\sin \alpha}{\alpha}\right)^2 \to 1$, leaving $I = I_0 \cos^2 \beta$, which is the standard interference formula.
Timed Topic Test
Attempt these questions under a ten-minute limit using a separate timer.
1. How many interference fringes lie between the central diffraction maximum and the first diffraction minimum on one side, given $d = 5a$?
Explanation: The ratio $d/a = 5$ indicates that the 5th interference maximum coincides with the 1st diffraction minimum, meaning there are 5 interference fringes within the half-envelope.
2. What is the physical reason behind 'missing orders' in a double-slit diffraction experiment?
Explanation: Missing orders occur because the single-slit diffraction envelope has a zero (minimum intensity) at the exact angle where the two-slit interference factor tries to create a maximum.
3. If the wavelength of light used in a double-slit diffraction setup is doubled, how does the angular width of the central diffraction maximum change?
Explanation: The angular width is $2\theta = \frac{2\lambda}{a}$, which is directly proportional to the wavelength $\lambda$.
4. In the intensity equation $I = I_0 \left(\frac{\sin \alpha}{\alpha}\right)^2 \cos^2 \beta$, what is the maximum value of the term $\left(\frac{\sin \alpha}{\alpha}\right)^2$?
Explanation: As $\alpha \to 0$ (at the central peak), $\lim_{\alpha \to 0} \left(\frac{\sin \alpha}{\alpha}\right)^2 = 1$.
5. If $d = 4\mu\text{m}$ and $a = 1\mu\text{m}$, which interference order will coincide with the first diffraction minimum ($m = 1$)?
Explanation: Using $\frac{d}{a} = \frac{n}{m}$, $\frac{4}{1} = \frac{n}{1}$, so $n = 4$ (the 4th interference order).
6. What happens to the intensity of the interference fringes as you move away from the central maximum towards higher diffraction angles?
Explanation: The fringes are modulated by the single-slit diffraction envelope, causing their peak intensities to decrease as $\alpha$ increases.
7. Which parameter solely controls the spacing between adjacent interference fringes in a double-slit diffraction pattern?
Explanation: The rapid interference fringe spacing depends on the interference condition governed by the slit separation $d$ (along with wavelength and screen distance).
8. If the slit separation $d$ is halved while keeping the slit width $a$ constant, what happens to the number of interference fringes within the central diffraction envelope?
Explanation: The number of fringes in the central envelope is $N = \frac{2d}{a}$. Halving $d$ halves $N$.
9. In double-slit diffraction, what condition must be satisfied for the 3rd interference order to be missing when $m = 1$?
Explanation: From $\frac{d}{a} = \frac{n}{m}$, setting $n = 3$ and $m = 1$ gives $\frac{d}{a} = 3$, meaning $d = 3a$.
10. Which approximation is typically assumed when stating that the diffraction minima occur at $a \sin \theta = m\lambda$?
Explanation: Fraunhofer diffraction assumes both the source and screen are effectively at infinity, leading to parallel rays and the standard single-slit formula.