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Basic Concepts of Diffraction
Diffraction is the bending of light waves around the corners of an obstacle or aperture into the geometrical shadow region. It arises due to the secondary wavelets originating from different points of the same wavefront, as stated by Huygens-Fresnel principle.
UP TGT ScienceUP PGT PhysicsStudy Notes + Practice
Scope and prerequisites
TGT core
Qualitative understanding of diffraction, distinction between Fresnel and Fraunhofer classes, condition for maxima and minima in single-slit diffraction, and basic wavelength dependence.
PGT extension
Analytical treatment of single-slit intensity distribution, phasor diagram method for resultant amplitude, width of central maximum, angular dispersion, and comparison with interference.
Prerequisites
Huygens' Principle and secondary wavelets
Superposition principle of wave motion
Phase difference and path difference relationships
Wavefront Subdivision and Huygens-Fresnel Principle
Every point on a propagating wavefront acts as a source of secondary spherical wavelets.
The resultant optical disturbance at any subsequent point is the superposition of all these secondary wavelets taking their respective phases and amplitudes into account.
Diffraction is fundamentally the interference of a continuous infinity of coherent secondary wavelets originating from the unobstructed portions of a single wavefront.
Classification of Diffraction: Fresnel and Fraunhofer
Fresnel diffraction occurs when either the source of light, the obstacle/aperture, or the screen (or any combination) is at a finite distance, making the wavefronts incident and emergent spherical or cylindrical.
Fraunhofer diffraction occurs when both the source and the observation screen are effectively at infinite distances from the aperture, achieved experimentally using lenses.
In Fraunhofer diffraction, the incident wavefronts are planar and the diffracted rays forming a particular parallel beam are brought to a focus on the screen.
Distinction Between Interference and Diffraction
Interference is the superposition of two or more distinct coherent wave trains originating from separate wavefronts (e.g., Young's double slit).
Diffraction is the superposition of continuous secondary wavelets originating from different parts of the same wavefront.
In interference, all bright fringes are generally of equal intensity, whereas in single-slit diffraction, the central maximum is brightest and widest, with subsidiary maxima falling off rapidly in intensity.
Conditions for Maxima and Minima in Single-Slit Diffraction
Secondary minima occur when the path difference between light rays originating from the top and bottom edges of a slit of width a is an integral multiple of the wavelength lambda.
Principal maxima occur approximately midway between the secondary minima, corresponding to odd multiples of lambda / 2 path difference for consecutive zones.
The central maximum is twice as wide as any of the secondary maxima and contains the vast majority of the incident optical energy.
Derivations
Fraunhofer Diffraction at a Single Slit
Conditions: Normal incidence of plane wavefront on a slit of width a, observation on a focal plane of a convex lens.
Consider a slit of width a divided into n equal strips. The path difference between rays from the extreme edges of the slit for a diffraction angle theta is given by delta = a sin(theta).
The phase difference between successive strips is bet$a = (2pi / \lambda) (a/n) \sin(\theta)$. Let the total phase difference between the extreme rays be 2alph$a = (2pi / \lambda) a \sin(\theta)$.
Summing the amplitudes of n secondary wavelets using the polygon method of vector addition yields the resultant amplitude $R = A0 (\sin(\alpha) / \alpha)$, where A0 is the maximum amplitude.
The intensity is proportional to R^2, giving $I = I0 (\sin(\alpha) / \alpha)^2$.
Result: $I = I_0 \left(\frac{\sin \alpha}{\alpha}\right)^2 \text{}$ where } alpha = \frac{pi a sin \theta}{lambda}
Position of Secondary Minima in Single-Slit Diffraction
Conditions: Destructive interference conditions across the slit width a.
Divide the slit width a into two equal halves of width a/2. For any ray from the upper half, a corresponding ray from the lower half can be found with a path difference of (lambda / 2).
This mutual cancellation leads to the first minimum when (a/2) sin $\theta = \lambda / 2$, or a sin \theta = lambda.
Generalizing this subdivision into 2n equal parts results in destructive interference whenever corresponding destructive pairs cancel out.
Therefore, the condition for the m-th secondary minimum is a sin \theta = mlambda, where m = pm 1, pm 2, pm 3, ...
Result: a sin \theta = mlambda quad (m = pm 1, pm 2, pm 3, ...)
Formula sheet
Formula / law
Equation
Variables and units
Conditions
Condition for Secondary Minima (Single Slit)
$$a \sin \theta = m\lambda$$
a = slit width, theta = diffraction angle, lambda = wavelength, $m = order integer (non-zero)$ a and lambda in meters, theta in radians
r$n = radius of n-th zone$, lambda = wavelength, a = source distance, b = screen distance Meters
Fresnel diffraction setup
Graphs and experiments
Intensity vs. Diffraction Parameter (alpha) Graph
Draw: A central symmetric peak of height I0 at alpha = 0, flanked by subsidiary secondary maxima of rapidly decreasing heights situated at alpha approx (m + 0.5)pi, with zero intensity at alph$a = m*\pi$.
Interpret: Demonstrates the rapid attenuation of intensity away from the central axis and highlights that the central maximum has twice the width of subsidiary maxima.
Exam use: Frequently used in PGT and TGT exams to test identification of minima positions and relative peak heights.
Fresnel's Biprisms or Half-Wave Plate / Zone Plate Setup
Draw: A circular aperture illuminated by a point source, showing concentric circular zones (Fresnel zones) of alternating phase cancellation.
Interpret: Illustrates how circular obstacles or apertures give rise to Fresnel diffraction patterns like the Poisson spot (Arago spot).
Exam use: Tested conceptually for differences between Fraunhofer and Fresnel diffraction geometries.
Single Slit Diffraction Experimental Setup
Draw: Monochromatic source, collimating lens L1, slit of width a, diffraction lens L2, and observation screen placed at the focal plane of L2.
Interpret: Shows how infinite distances are simulated in a laboratory bench using lenses for Fraunhofer diffraction observations.
Exam use: Conceptual layout questions in practical optics and instrumentation.
Solved numericals
TGT core
Light of wavelength 5000 Å falls normally on a single slit of width 0.2 mm. Calculate the angular position of the first secondary minimum.
Identify given parameters: lambd$a = 5000 Å = 5 x 10^-7 m$, $a = 0.2 mm = 2 x 10^-4 m$, m = 1.
Use the diffraction minimum condition: a sin(theta) = m * lambda.
Substitute values: sin(theta) = (1 * 5 x 10^-7) / (2 x 10^-4) = 2.5 x 10^-3.
Since sin(theta) is very small, theta approx sin(theta) = 2.5 x 10^-3 radians.
Answer: 2.5 x 10^-3 rad
PGT extension
A slit of width 0.5 mm is illuminated normally by light of wavelength 589 nm. The diffraction pattern is observed on a screen placed 1.0 m away from the lens. Calculate the linear width of the central maximum.
Identify given data: $a = 0.5 mm = 5 x 10^-4 m$, lambd$a = 589 nm = 5.89 x 10^-7 m$, f (or D) = 1.0 m.
Recall the formula for linear width of central maximum: bet$a_0 = 2 * \lambda * f / a$.
Substitute the values: bet$a_0 = (2 * 5.89 x 10^-7 * 1.0) / (5 x 10^-4)$.
Calculate the final numerical value: bet$a_0 = (11.78 x 10^-7) / (5 x 10^-4) = 2.356 x 10^-3 m$.
Answer: 2.356 mm
PGT extension
In a single slit diffraction pattern, the angular width of the central maximum is measured to be 2 degrees when light of wavelength 600 nm is used. Find the width of the slit.
Identify the angular width relation: Total angular widt$h = 2 * \theta = 2 * (\lambda / a)$.
Use thet$a = \lambda / a$ to find a: $a = \lambda / \theta$.
Substitute lambd$a = 600 x 10^-9 m$ and thet$a = (\pi / 180) rad: a = (600 x 10^-9 * 180) / \pi$.
Evaluate: $a = (1.08 x 10^-4) / 3.1416 \approx 3.438 x 10^-5 m = 0.0344 mm$.
Answer: 3.44 x 10^-5 m (or 0.0344 mm)
Assumptions and limitations
Scalar diffraction theory assumes the light wave can be treated as a scalar wave, ignoring vector polarization effects, which holds well when the aperture dimensions are much larger than the wavelength.
Fraunhofer approximations assume that distances R >> a^2 / lambda, validating the neglect of quadratic phase terms across the aperture.
Common misconceptions
Common error: All bright fringes in a single-slit diffraction pattern have equal intensity.
Correct: Only the central maximum has maximum intensity; subsidiary maxima have drastically lower intensities.
Unlike interference fringes where amplitudes from two coherent slits remain uniform, diffraction involves varying fractional cancellation across a continuous aperture.
Common error: Diffraction only occurs when obstacles are microscopic.
Correct: Diffraction occurs for obstacles of any size, but it is pronounced only when the obstacle or aperture dimensions are comparable to the wavelength of the wave.
The diffraction angle theta approx lambda / a is too small to detect easily when aperture a is extremely large compared to lambda.
Quick revision
Diffraction is the bending of waves around edges governed by secondary wavelets from unblocked wavefronts.
Fraunhofer diffraction requires infinite source-screen distances, simulated in labs via lenses.
Minima condition for single slit: a sin(theta) = m*lambda (m != 0).
Central maximum width is twice that of any secondary maximum: bet$a_0 = 2*\lambda*f / a$.
Must remember
Diffraction is fundamentally the interference of a continuous infinity of coherent secondary wavelets originating from the unobstructed portions of a single wavefront.
Fresnel diffraction involves finite distances for source and/or screen, producing spherical or cylindrical wavefronts.
Fraunhofer diffraction requires both source and screen to be effectively at infinite distances, realized using lenses to handle planar wavefronts.
The condition for secondary minima in single-slit diffraction is $a \sin \theta = m\lambda$, where $m = \pm 1, \pm 2, \dots$.
The linear width of the central maximum on a screen placed at the focal plane of a converging lens is given by $\beta_0 = \frac{2f\lambda}{a}$.
Unlike interference fringes, single-slit diffraction subsidiary maxima fall off rapidly in intensity and the central maximum is twice as wide.
Glossary
Fresnel DiffractionDiffraction phenomenon where source, obstacle, or screen are at finite distances, resulting in spherical or cylindrical wavefronts.
Fraunhofer DiffractionDiffraction phenomenon where both source and observation screen are effectively at infinite distances, with planar incident and emergent wavefronts.
Slit Width ($a$)The aperture dimension through which light passes in single-slit diffraction experiments.
Half-Angular WidthThe angular spread from the center of the diffraction pattern to the first secondary minimum, approximated as $\theta \approx \frac{\lambda}{a}$.
Interference vs. Diffraction
Parameter
Interference
Diffraction
Origin
Superposition of two or more distinct coherent wave trains from separate wavefronts
Superposition of continuous secondary wavelets from different parts of the same wavefront
Intensity of Bright Fringes
All bright fringes are generally of equal intensity
Central maximum is brightest and widest; subsidiary maxima fall off rapidly in intensity
Width of Fringes
Fringes are typically of equal width
Central maximum is twice as wide as any subsidiary maximum
Practice Quiz
Choose an option to reveal the explanation.
1. What is the primary physical origin of diffraction according to the Huygens-Fresnel principle?
Explanation: Diffraction is fundamentally the interference of a continuous infinity of coherent secondary wavelets originating from the unobstructed portions of a single wavefront.
2. In which type of diffraction are the incident and emergent wavefronts planar?
Explanation: Fraunhofer diffraction occurs when both the source and the observation screen are effectively at infinite distances, so the incident wavefronts are planar.
3. What is the condition for secondary minima in single-slit diffraction of width $a$ for wavelength $\lambda$?
Explanation: Secondary minima occur when the path difference between extreme rays is an integral multiple of wavelength, given by $a \sin \theta = m\lambda$ ($m \neq 0$).
4. How does the intensity of subsidiary maxima compare to the central maximum in single-slit diffraction?
Explanation: In single-slit diffraction, the central maximum is brightest and widest, with subsidiary maxima falling off rapidly in intensity.
5. What is the linear width of the central maximum $\beta_0$ on a screen placed at the focal plane of a lens of focal length $f$?
Explanation: The linear width of the central maximum is $\beta_0 = \frac{2f\lambda}{a}$, where $a$ is the slit width.
6. Which diffraction class is characterized by finite distances between the source, aperture, and screen?
Explanation: Fresnel diffraction occurs when either the source, obstacle/aperture, or screen is at a finite distance, yielding spherical or cylindrical wavefronts.
7. What is the half-angular width of the central maximum for a narrow slit of width $a$ illuminated by light of wavelength $\lambda$?
Explanation: Using the small angle approximation for narrow slits, $\sin \theta \approx \theta = \frac{\lambda}{a}$.
8. Which of the following correctly defines the phase difference parameter $\alpha$ in single-slit diffraction?
Explanation: The phase difference parameter is defined as $\alpha = \frac{\pi a \sin \theta}{\lambda}$, representing the half-phase difference between extreme rays.
9. What is the intensity distribution formula for Fraunhofer single-slit diffraction?
Explanation: The intensity distribution is given by $I = I_0 \left(\frac{\sin \alpha}{\alpha}\right)^2$, where $\alpha = \frac{\pi a \sin \theta}{\lambda}$.
10. Why is diffraction pronounced only when obstacle dimensions are comparable to the wavelength of the wave?
Explanation: When aperture $a$ is extremely large compared to $\lambda$, the diffraction angle $\theta \approx \frac{\lambda}{a}$ is vanishingly small, making the effect undetectable.
11. What is the path difference between extreme rays originating from the top and bottom edges of a slit of width $a$ at diffraction angle $\theta$?
Explanation: The path difference between extreme rays for normal incidence is given by $\Delta = a \sin \theta$.
12. In interference (e.g., Young's double slit), how do bright fringes compare in intensity?
Explanation: In interference, all bright fringes are generally of equal intensity, unlike single-slit diffraction where subsidiary maxima fall off rapidly.
13. What experimental arrangement is typically used to achieve Fraunhofer diffraction in a laboratory?
Explanation: Fraunhofer diffraction is achieved experimentally using lenses so that both the source and the observation screen are effectively at infinite distances.
14. What is the approximate condition for secondary maxima (excluding the central maximum) in single-slit diffraction?
Explanation: Principal/secondary maxima occur approximately midway between secondary minima, corresponding to odd multiples of $\frac{\lambda}{2}$, i.e., $a \sin \theta = \left(m + \frac{1}{2}\right)\lambda$.
15. Which of the following is a common misconception about single-slit diffraction?
Explanation: It is a common misconception that all bright fringes have equal intensity; in reality, subsidiary maxima drop off drastically in intensity compared to the central maximum.
PYQ Patterns
These are pattern-based practice questions. Authentic year labels are only used when verified from a source.
1. A slit of width $0.2 \text{ mm}$ is illuminated by monochromatic light of wavelength $5890 \text{ \AA}$. What is the angular half-width of the central maximum?
2. In a Fraunhofer diffraction pattern by a single slit, the first secondary minimum is observed at an angle of $30^\circ$ for a light of wavelength $5000 \text{ \AA}$. What is the width of the slit?
Explanation: For the first secondary minimum ($m=1$), $a \sin \theta = 1\lambda \implies a \sin(30^\circ) = 5000 \times 10^{-10} \implies a(0.5) = 5 \times 10^{-7} \implies a = 1 \times 10^{-6} \text{ m}.
3. What is the ratio of the width of the central maximum to that of any secondary maximum in single-slit diffraction?
Explanation: The angular width of the central maximum is $2\frac{\lambda}{a}$, while the angular width of any subsidiary maximum is $\frac{\lambda}{a}$. Thus, the ratio is $2:1$.
4. If the slit width in a single-slit diffraction experiment is halved, how does the angular width of the central maximum change?
Explanation: Since angular half-width $\theta = \frac{\lambda}{a}$, halving $a$ doubles the angular width of the central maximum.
5. Light of wavelength $6000 \text{ \AA}$ falls normally on a slit. If the first minimum is formed at $\theta = 30^\circ$, calculate the slit width $a$.
Explanation: Using $a \sin \theta = \lambda$ for $m=1$, $a \sin(30^\circ) = 6000 \times 10^{-10} \implies a(0.5) = 6 \times 10^{-7} \implies a = 1.2 \times 10^{-6} \text{ m}.
6. Which of the following statements correctly distinguishes diffraction from interference?
Explanation: Interference is the superposition of distinct coherent wave trains from separate wavefronts, while diffraction is the superposition of continuous secondary wavelets from different parts of the same wavefront.
Timed Topic Test
Attempt these questions under a ten-minute limit using a separate timer.
1. What happens to the diffraction pattern of a single slit if the monochromatic light source is replaced by a white light source?
Explanation: With white light, the central maximum for all wavelengths coincides at $\theta = 0$ forming a white fringe, while subsidiary maxima are dispersed into colored spectra because their positions depend on $\lambda$.
2. In Fraunhofer diffraction by a single slit, the intensity of the first subsidiary maximum is approximately what fraction of the central maximum intensity $I_0$?
Explanation: The first subsidiary maximum occurs at $\alpha \approx \frac{3\pi}{2}$. Substituting into $I = I_0 \left(\frac{\sin \alpha}{\alpha}\right)^2$ gives $I \approx I_0 \left(\frac{1}{3\pi/2}\right)^2 = \frac{4}{9\pi^2} I_0$ (roughly $4.5\%$ of $I_0$).
3. What is the radius of the $n$-th Fresnel zone for a point source and screen setup given by $r_n = \sqrt{\frac{n \lambda b}{1 + b/a}}$? What does $b$ represent?
Explanation: In the Fresnel zone expression, $a$ represents the source distance and $b$ represents the screen (observation point) distance from the aperture.
4. If the wavelength of light used in a single-slit diffraction setup is doubled while keeping the slit width constant, what happens to the linear width of the central maximum?
Explanation: Since $\beta_0 = \frac{2f\lambda}{a}$, the linear width is directly proportional to wavelength $\lambda$. Doubling $\lambda$ doubles the linear width.
5. Which condition separates Fresnel diffraction from Fraunhofer diffraction?
Explanation: Fresnel diffraction occurs at finite distances (spherical/cylindrical wavefronts), whereas Fraunhofer diffraction occurs at infinite distances (planar wavefronts, achieved via lenses).
6. In single-slit diffraction, what occurs at angles where $\alpha$ is an integral multiple of $\pi$ (excluding zero)?
Explanation: When $\alpha = m\pi$ ($m \neq 0$), $\sin \alpha = 0$, which corresponds to secondary minima and zero optical intensity.
7. Why can sound waves bend around corners of a doorway more noticeably than light waves?
Explanation: Diffraction is prominent when obstacle dimensions are comparable to wavelength. Sound wavelengths are in meters (comparable to doors), whereas light wavelengths are in nanometers.
8. What is the effect on the diffraction pattern if a transparent thin film is introduced in front of one half of the single slit?
Explanation: Introducing a phase-changing thin film across part of the slit introduces an additional optical path difference, causing a lateral shift of the entire diffraction pattern on the screen.
9. What is the phase difference between wavelets originating from the top and bottom edges of a single slit when the path difference is equal to $\lambda$?
Explanation: A path difference of $\lambda$ corresponds to a phase difference of $2\pi$ radians between extreme rays across the entire slit width $a$.
10. Which of the following parameters does NOT directly affect the angular position of secondary minima in single-slit Fraunhofer diffraction?
Explanation: The angular positions are given by $a \sin \theta = m\lambda$, which depends on $a$, $\lambda$, and $m$. The focal length $f$ affects linear positions on the focal plane, not angular positions.