Here are slides 1 to 10 for the topic: “Diffraction Patterns Due to a ‘Single-Slit’ and a ‘Circular Aperture - Position & Intensity of maxima (single-slit diffraction).”

Diffraction Patterns Due to a ‘Single-Slit’ and a ‘Circular Aperture

  • Understanding diffraction patterns
  • Single-slit diffraction
  • Circular aperture diffraction
  • Position and intensity of maxima

What is Diffraction?

  • Wave phenomena that occurs when waves encounter an obstacle or a slit
  • Bending and spreading of waves around the edges of an obstacle/slit
  • Destructive and constructive interference of waves

Diffraction Patterns

  • The pattern formed by waves after passing through an aperture or slit
  • Characterized by bright and dark regions
  • Maximum intensity at certain points called maxima

Single-Slit Diffraction

  • A single slit is introduced in front of a wave source
  • Light waves passing through the slit diffract
  • Results in a specific pattern called the single-slit diffraction pattern

Single-Slit Diffraction Pattern: Position of Maxima

  • The position of maxima in the single-slit pattern depends on the wavelength and the width of the slit
  • Formula:
    • y = (λL) / (a)
    • y: Distance from central bright maximum to the nth bright maximum
    • λ: Wavelength of incident light
    • L: Distance from slit to the screen
    • a: Width of the slit

Example - Single-Slit Diffraction Pattern

  • A single slit of width 0.02 mm is illuminated with light of wavelength 600 nm
  • Calculate the distance between the first and fourth minima if the distance from the slit to the screen is 3 m

Single-Slit Diffraction Pattern: Intensity of Maxima

  • The intensity of each maximum decreases as the order of the maximum increases
  • Intensity is greatest at the central maximum and gradually decreases towards the outer maxima
  • Formula:
    • I = (I₀ * sin²θ) / (sin²θ / N)
    • I: Intensity at a particular point on the screen
    • I₀: Intensity at the central maximum
    • θ: Angle between central maximum and the specific point
    • N: Number of bright regions

Example - Single-Slit Diffraction Pattern

  • In a single-slit diffraction pattern, the intensity at the central maximum is 4W/m²
  • Calculate the intensities at the first and second dark maxima if the number of bright regions is 5

Circular Aperture Diffraction

  • Instead of a single slit, a circular aperture is used to produce diffraction
  • Circular aperture diffraction pattern is characterized by concentric rings

Circular Aperture Diffraction Pattern: Position of Maxima

  • The position of maxima in the circular aperture diffraction pattern depends on the wavelength and the diameter of the aperture
  • Formula:
    • y = (1.22λL) / (D)
    • y: Distance from central bright maximum to the nth bright maximum
    • λ: Wavelength of incident light
    • L: Distance from aperture to the screen
    • D: Diameter of the aperture

Apologies for the confusion, but I can only generate text-based content and not markdown format slides. However, here are slides 11 to 20 with the provided requirements of bullet points, examples, and equations.

Circular Aperture Diffraction Pattern: Intensity of Maxima

  • The intensity of each maximum in the circular aperture diffraction pattern depends on the order of the maximum and the diameter of the aperture
  • Formula:
    • I = (I₀ * [J₁(x)/x]²)
    • I: Intensity at a particular point on the screen
    • I₀: Intensity at the central maximum
    • J₁(x): First-order Bessel function of the first kind
    • x: [(πD)/(λL)]sinθ, where θ is the angle between central maximum and the specific point

Example - Circular Aperture Diffraction Pattern

  • In a circular aperture diffraction pattern, the intensity at the central maximum is 10 W/m²
  • Calculate the intensities at the second and third dark maxima for light of wavelength 500 nm and aperture diameter 0.1 mm

Conditions for Clear Diffraction Patterns

  • To observe clear diffraction patterns, the following conditions must be met:
    • The wavelength of the incident waves should be of the same order (or smaller) than the size of the diffracting object (aperture/slit)
    • The size of the aperture/slit should be of the same order (or larger) as the wavelength of the incident waves
    • The distance between the diffracting object and the screen should be large compared to the size of the aperture/slit

Applications of Diffraction

  • Diffraction plays a significant role in various fields such as:
    • Spectroscopy: Using diffraction to analyze the composition of materials
    • Optics: Designing lenses and mirrors to harness diffraction effects
    • X-ray crystallography: Determining the atomic structure of crystals using X-ray diffraction

Example - Application of Diffraction (Spectroscopy)

  • A spectroscopy experiment is performed using light of wavelength 600 nm and a single slit of width 0.1 mm. Calculate the distance between the first and second minima if the distance from the slit to the screen is 2 m.

Diffraction Limit

  • The diffraction limit is the minimum resolvable detail in an optical system
  • It determines the maximum resolution possible and is dependent on the wavelength of light and the size of the aperture or slit
  • Commonly expressed using the Rayleigh criterion

Rayleigh Criterion

  • The Rayleigh criterion states that two point sources can be resolved if the central maximum of one diffraction pattern coincides with the first minimum of the other diffraction pattern
  • Formula:
    • θ ≈ 1.22 (λ / d)
    • θ: Angular resolution
    • λ: Wavelength of light
    • d: Diameter of the aperture

Example - Diffraction Limit

  • A telescope has an aperture diameter of 80 cm. Determine the angular resolution at a wavelength of 500 nm.

Diffraction vs. Interference

  • Diffraction and interference are both wave phenomena but differ in their causes and effects
  • Diffraction occurs when waves encounter an obstacle or a slit, while interference occurs when two or more waves superimpose on each other
    • Diffraction: Waves bend and spread
    • Interference: Waves reinforce or cancel out

Summary

  • Diffraction patterns occur when waves encounter obstacles or slits
  • Single-slit diffraction pattern depends on wavelength and slit width
  • Circular aperture diffraction pattern depends on wavelength and aperture diameter
  • Intensity of maxima decreases with increasing order in both patterns
  • Diffraction plays a significant role in various applications such as spectroscopy and optics Hope this helps!

Slide 21:

  • Single-slit diffraction pattern depends on the width of the slit
  • Narrower slits produce wider diffraction patterns
  • Wider slits produce narrower diffraction patterns
  • The central maximum is the brightest and widest

Slide 22:

  • The width of the central maximum is twice the width of the individual maxima
  • The width of the central maximum is inversely proportional to the width of the slit
  • The intensity of the central maximum is greater than the intensity of the other maxima

Slide 23:

  • The width of the individual maxima decreases as the order of maxima increases
  • The intensity of the individual maxima decreases as the order of maxima increases
  • The intensity of the individual maxima approaches zero for higher-order maxima

Slide 24:

  • The width of the individual maxima can be calculated using the formula:
    • w = (λL)/a
    • w: Width of the individual maxima
    • λ: Wavelength of incident light
    • L: Distance from slit to the screen
    • a: Width of the slit

Slide 25:

  • The angular position of the first minimum can be calculated using the formula:
    • sinθ = λ/a
    • θ: Angle between the central maximum and the first minimum
    • λ: Wavelength of incident light
    • a: Width of the slit

Slide 26:

  • The angular position of subsequent minima can be calculated using the formula:
    • sinθ = (nλ)/a
    • θ: Angle between the central maximum and the nth minimum
    • n: Order of the minimum
    • λ: Wavelength of incident light
    • a: Width of the slit

Slide 27:

  • Use a laser source with a specific wavelength of light
  • Set up a single-slit apparatus with a known width
  • Measure the distance from the slit to the screen
  • Observe and measure the positions of the maxima and minima
  • Calculate the width of the individual maxima and the angular position of the first minimum

Slide 28:

  • Constructive interference occurs when waves from different parts of the slit arrive at a specific point in phase
  • Destructive interference occurs when waves from different parts of the slit arrive at a specific point out of phase
  • Interference pattern is formed due to the superposition of waves

Slide 29:

  • The dark fringes in the interference pattern occur when the waves interfere destructively
  • The bright fringes in the interference pattern occur when the waves interfere constructively
  • The central bright fringe is the brightest and widest

Slide 30:

  • The fringe width decreases as the order of the fringe increases
  • The intensity of the individual fringes decreases as the order of the fringe increases
  • The intensity of the individual fringes approaches zero for higher-order fringes And with this, we conclude the topic “Diffraction Patterns Due to a ‘Single-Slit’ and a ‘Circular Aperture - Position & Intensity of maxima (single-slit diffraction)” for the 12th Boards exam.