Diffraction Comprehensive Notes

Diffraction

Introduction

  • Diffraction is a phenomenon where waves, including sound and light, bend around corners or obstructions.

  • This explains why we can hear someone talking even when they are around a corner.

  • The extent of diffraction depends on the wavelength of the wave and the size of the obstruction or opening.

  • Longer wavelengths diffract more than shorter wavelengths.

  • Sound waves (wavelength ~1 meter) diffract more easily than light waves (wavelength ~ 10^{-7} meters).

Diffraction of Water Waves

  • When water waves pass through an opening, they spread outward, or diffract.

  • An observer at point P detects waves even though this point is not on a line with the original direction of the waves through the opening.

  • All waves exhibit similar behavior.

Single-Slit Diffraction

  • Single-slit diffraction occurs when light passes through a tall, narrow slit of width a.

  • The resulting pattern on a viewing screen consists of a central maximum (brightest and broadest) and a series of weaker secondary maxima and dark fringes.

Analyzing Single-Slit Diffraction

  • The bright central maximum is located at θ = 0° and has the highest intensity.

  • Dark fringes (minima) occur at specific angles determined by the condition for destructive interference.

Conditions for Dark Fringes in Single-Slit Interference

  • Destructive interference occurs when the path-length difference between waves from different parts of the slit is an integer multiple of the wavelength.

  • The condition for dark fringes is given by: a sin θ = mλ, where:

    • a is the width of the slit.

    • θ is the angle to the dark fringe.

    • m is an integer (\pm 1, \pm 2, \pm 3,…) representing the order of the dark fringe.

    • \lambda is the wavelength of the light.

Small Angle Approximation

  • Using small angle approximation, position of the minima: y_m = \frac{mλL}{a}

Width of the Central Bright Fringe

  • The width w of the central maximum is defined as the distance between the two m=1 minima.

  • w = 2y_1 = \frac{2λL}{a}

  • Where:

    • L is the distance from the slit to the screen.

Example Calculation

  • Monochromatic light passes through a slit of width a = 1.2 × 10^{-5} m.

  • The first dark fringe is observed at an angle of θ = 3.25°.

  • To find the wavelength of the light, use the formula: \lambda = \frac{a sin θ}{m}

  • \lambda = \frac{(1.2 × 10^{-5} m) sin (3.25°)}{1} = 6.80 × 10^{-7} m

Conceptual Understanding

  • If the wavelength of the light is increased, the angle to the first dark fringe also increases.

  • The smaller the opening a wave squeezes through, the more it spreads out on the other side.

Diffraction Grating

  • A diffraction grating consists of a number of equally spaced parallel slits.

  • It is used for analyzing light sources by separating light into its component wavelengths.

Slit Spacing

  • The slit spacing d for a diffraction grating with n lines/cm is given by: d = \frac{1}{n}

Condition for Maxima in Interference Pattern

  • The condition for maxima (bright fringes) in the interference pattern at an angle θ is: d sin θ_{bright} = mλ, where:

    • d is the slit spacing.

    • θ_{bright} is the angle to the bright fringe.

    • m is an integer (0, \pm 1, \pm 2,…) representing the order of the maximum.

    • \lambda is the wavelength of the light.

Effect of Wavelength on Diffraction Patterns

  • Different wavelengths of light are diffracted at different angles, resulting in a spectrum.

Calculating Principal Maxima

  • To calculate different-order principal maxima for a diffraction grating, use the formula: d sin θ = mλ

Example 24.7: A Diffraction Grating

  • Monochromatic light from a helium-neon laser (\lambda = 632.8 nm) is incident on a diffraction grating with 6.00 × 10^3 lines/cm.

  • First, find the slit separation: d = \frac{1}{6.00 × 10^3 cm^{-1}} = 1.67 × 10^{-4} cm = 1.67 × 10^3 nm.

  • For the first-order maximum (m = 1): sin θ1 = \frac{λ}{d} = \frac{632.8 nm}{1.67 × 10^3 nm} = 0.379, so θ1 = sin^{-1}(0.379) = 22.3°.

  • For the second-order maximum (m = 2): sin θ2 = \frac{2λ}{d} = \frac{2(632.8 nm)}{1.67 × 10^3 nm} = 0.758, so θ2 = 49.3°.

  • For the third-order maximum (m = 3): sin θ3 = \frac{3λ}{d} = \frac{3(632.8 nm)}{1.67 × 10^3 nm} = 1.14. Since sin θ cannot exceed 1, there is no solution for θ3.

Diffraction Grating Equation

  • d sin(θ) = nλ

  • Where:

    • d = Slit spacing

    • θ = Angle of diffraction

    • n = Order of the maximum

    • λ = Wavelength of light

Worked Example

  • An experiment uses a diffraction grating with a slit spacing of 1.7 μm, and the wavelength of light is 550 nm.

  • To find the angle θ for the second-order line (n=2):

  • Rearrange the formula: sin(θ) = \frac{nλ}{d}

  • Substitute the values: sin(θ) = \frac{2 × 550 × 10^{-9}}{1.7 × 10^{-6}} = 0.64705… ≈ 0.65

  • Find the angle: θ = sin^{-1}(0.65) = 40.54°

  • The angle between the two second-order lines is 2 × 40.54° ≈ 81°.