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): $sin θ<em>1 = \frac{λ}{d} = \frac{632.8 nm}{1.67 × 10^3 nm} = 0.379$, so $θ</em>1 = sin^{-1}(0.379) = 22.3°$.

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

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

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 $2 × 40.54° ≈ 81°$.