Slide 1

  • Topic: Diffraction - Intensity Distribution on the focal plane
  • Definition: Diffraction is the bending or spreading out of a wave when it encounters an obstacle or passes through a narrow slit.
  • Diffraction can occur with any type of wave, including light waves, sound waves, and water waves.
  • In this lecture, we will focus on the diffraction of light waves.
  • One important phenomenon related to diffraction is the intensity distribution on the focal plane.

Slide 2

  • Diffraction Grating:
    • A diffraction grating is a device that consists of a large number of equally spaced parallel slits or lines.
    • It is used to study the diffraction of light and analyze its properties.
    • The spacing between the slits or lines in a diffraction grating is typically on the order of the wavelength of light.
  • When light passes through a diffraction grating:
    • Each slit acts as a source of circular wavelets.
    • These wavelets interfere and produce constructive and destructive interference.
    • The resulting interference pattern can be observed on a screen placed on the focal plane of the grating.

Slide 3

  • Constructive Interference:
    • When the path difference between two adjacent wavelets is an integral multiple of the wavelength, they interfere constructively.
    • This results in bright fringes in the interference pattern.
    • The bright fringes are characterized by high intensity.
  • Destructive Interference:
    • When the path difference between two adjacent wavelets is a half-integral multiple of the wavelength, they interfere destructively.
    • This results in dark fringes in the interference pattern.
    • The dark fringes are characterized by low or zero intensity.

Slide 4

  • Equation for Intensity Distribution:
    • The intensity of the interference pattern on the focal plane of a diffraction grating can be calculated using the equation:
      • I = I₀ * [sin(α)/α]²
    • In this equation:
      • I is the intensity at a particular point on the focal plane,
      • I₀ is the intensity of the central maximum,
      • α is the angle between the direction of the incident light and the normal to the grating.

Slide 5

  • Example: Light Passing through a Diffraction Grating
    • Consider a diffraction grating with a separation between slits or lines of 1.0 × 10⁻⁵ meters.
    • A beam of monochromatic light with a wavelength of 5.0 × 10⁻⁷ meters is incident on the grating.
    • Find the angle at which the first-order bright fringe will appear on the focal plane.
  • Solution:
    • Given: λ = 5.0 × 10⁻⁷ meters, d = 1.0 × 10⁻⁵ meters.
    • Using the equation for constructive interference: d * sin(θ) = m * λ
    • For the first-order bright fringe: m = 1
    • Substituting the values: 1.0 × 10⁻⁵ * sin(θ) = 1 * 5.0 × 10⁻⁷
    • Solving for θ: sin(θ) = 5.0 × 10⁻²
    • Taking the inverse sine: θ = sin⁻¹(5.0 × 10⁻²)

Slide 6

  • Example (continued):
    • Evaluating the value using a scientific calculator: θ ≈ 2.87°
  • Therefore, the angle at which the first-order bright fringe will appear on the focal plane is approximately 2.87°.

Slide 7

  • Number of Fringes:
    • The number of bright or dark fringes that can be observed on the focal plane depends on the order of the fringe and the aperture size of the grating.
    • Higher-order fringes occur farther away from the central maximum.
    • As the order increases, the fringes become more closely spaced.
  • The formula to calculate the number of fringes:
    • N = (m₁ - m₂) * d / λ
    • In this equation:
      • N is the number of fringes,
      • m₁ and m₂ are the orders of the fringes,
      • d is the slit or line separation,
      • λ is the wavelength of light.

Slide 8

  • Example: Calculation of Number of Fringes
    • A diffraction grating with a slit separation of 2.0 × 10⁻⁶ meters is used to study the interference pattern of a monochromatic light source. The wavelength of light is 5.0 × 10⁻⁷ meters. Calculate the number of fringes between the third-order bright fringe and the fourth-order dark fringe.
  • Solution:
    • Given: d = 2.0 × 10⁻⁶ meters, λ = 5.0 × 10⁻⁷ meters, m₁ = 3, m₂ = 4.
    • Using the formula for calculating the number of fringes: N = (m₁ - m₂) * d / λ
    • Substituting the given values: N = (3 - 4) * 2.0 × 10⁻⁶ / 5.0 × 10⁻⁷
    • Simplifying the expression: N = -2 * 2.0 × 10⁻⁶ / 5.0 × 10⁻⁷

Slide 9

  • Example (continued):
    • Evaluating the expression: N = -20.0
  • Therefore, the number of fringes between the third-order bright fringe and the fourth-order dark fringe is -20.
  • Note: The negative sign indicates that the fringe order is decreasing from m₁ to m₂.

Slide 10

  • Conclusion:
    • Diffraction is the bending or spreading out of a wave when it encounters an obstacle or passes through a narrow slit.
    • Diffraction grating is a device consisting of parallel slits or lines used to study the diffraction of light.
    • Intensity distribution on the focal plane of a diffraction grating is characterized by bright and dark fringes.
    • The intensity distribution can be calculated using the equation I = I₀ * [sin(α)/α]².
    • The number of fringes between different orders can be determined using the formula N = (m₁ - m₂) * d / λ.

Slide 11

  • Diffraction Grating Equation:
    • The diffraction grating equation relates the wavelength of light to the angle at which the bright fringes occur on the focal plane.
    • The equation is given by: d * sin(θ) = m * λ
      • d is the separation between the slits or lines of the grating,
      • θ is the angle at which the bright fringe occurs,
      • m is the order of the bright fringe,
      • λ is the wavelength of light.
  • The diffraction grating equation is derived based on the principle of constructive interference in the grating.
  • The equation allows us to calculate the wavelength of light using the known values of the slit separation and the angle of diffraction.

Slide 12

  • Example: Calculation of Wavelength using Diffraction Grating Equation
    • A diffraction grating with a slit separation of 5.0 × 10⁻⁶ meters produces a first-order bright fringe at an angle of 30°. Calculate the wavelength of light used.
  • Solution:
    • Given: d = 5.0 × 10⁻⁶ meters, θ = 30°, m = 1.
    • Using the diffraction grating equation: d * sin(θ) = m * λ
    • Substituting the given values: 5.0 × 10⁻⁶ * sin(30°) = 1 * λ
    • Simplifying the expression: 5.0 × 10⁻⁶ * 0.5 = λ
    • Evaluating the expression: λ = 2.5 × 10⁻⁶ meters
  • Therefore, the wavelength of light used is approximately 2.5 × 10⁻⁶ meters.

Slide 13

  • Limitations of Diffraction Grating:
    • The diffraction grating equation is based on the assumption of planar wavefronts.
    • It assumes that the light waves incident on the grating are parallel and have a uniform intensity across the wavefront.
    • In reality, light waves diverge from a point source and have variations in intensity across the wavefront.
    • These deviations from the ideal conditions can affect the accuracy of calculations using the diffraction grating equation.
  • Despite its limitations, the diffraction grating equation is still widely used in experimental setups and practical applications.

Slide 14

  • Resolving Power of Diffraction Grating:
    • The resolving power of a diffraction grating is a measure of its ability to distinguish between closely spaced spectral lines of different wavelengths.
    • It is given by the equation: R = N * m
      • R is the resolving power,
      • N is the number of slits or lines per unit length,
      • m is the order of the spectral line.
  • The resolving power of a diffraction grating increases with:
    • Higher order of the spectral line,
    • More slits or lines per unit length.

Slide 15

  • Example: Resolving Power of a Diffraction Grating
    • A diffraction grating has 5000 lines per centimeter and is used to separate the red (wavelength = 700 nm) and blue (wavelength = 450 nm) light. Calculate the resolving power of the grating for the first-order spectral lines.
  • Solution:
    • Given: N = 5000 lines/cm, λ₁ = 700 nm, λ₂ = 450 nm, m = 1.
    • Using the formula for resolving power: R = N * m
    • Substituting the given values: R = 5000 * 1
    • Evaluating the expression: R = 5000
  • Therefore, the resolving power of the diffraction grating is 5000 for the first-order spectral lines.

Slide 16

  • Spectral Dispersion:
    • Spectral dispersion refers to the separation of different wavelengths of light by a diffraction grating.
    • It is due to the phenomenon of diffraction and the interference of wavelets from different slits or lines.
    • A diffraction grating produces a series of spectral lines separated by angular or linear distances.
  • Spectral dispersion is useful in spectroscopy and other scientific applications where the analysis of light at different wavelengths is required.

Slide 17

  • Example: Spectral Dispersion of a Diffraction Grating
    • A diffraction grating with a slit separation of 2.0 × 10⁻⁶ meters is used to analyze a light source. Calculate the angular separation between the first-order spectral lines for two different wavelengths of light: 600 nm and 800 nm.
  • Solution:
    • Given: d = 2.0 × 10⁻⁶ meters, λ₁ = 600 nm, λ₂ = 800 nm, m = 1.
    • Using the diffraction grating equation: d * sin(θ₁) = m * λ₁ and d * sin(θ₂) = m * λ₂
    • Dividing the two equations: sin(θ₁) / sin(θ₂) = λ₁ / λ₂
    • Substituting the given values: sin(θ₁) / sin(θ₂) = 600 / 800
    • Evaluating the expression using a scientific calculator: θ₁ / θ₂ ≈ 0.75

Slide 18

  • Example (continued):
    • Taking the inverse sine: θ₁ ≈ sin⁻¹(0.75)
    • Evaluating the value using a scientific calculator: θ₁ ≈ 48.6°
  • Therefore, the angular separation between the first-order spectral lines for the wavelengths 600 nm and 800 nm is approximately 48.6°.

Slide 19

  • Applications of Diffraction Grating:
    • Diffraction gratings are widely used in various areas of science and technology.
    • They are used in spectrometers, spectroscopes, and other optical instruments for analyzing the composition of light sources.
    • Diffraction gratings are also used in telecommunications to separate different wavelengths of light in fiber optic systems.
    • They find applications in laser display systems, holography, and astronomy for studying the spectra of celestial objects.

Slide 20

  • Conclusion:
    • The diffraction grating equation relates the wavelength of light to the angle at which the bright fringes occur on the focal plane.
    • The resolving power of a diffraction grating is a measure of its ability to distinguish between closely spaced spectral lines of different wavelengths.
    • Spectral dispersion refers to the separation of different wavelengths of light by a diffraction grating.
    • Diffraction gratings have various applications in spectroscopy, telecommunications, display systems, holography, and astronomy.

Slide 21

  • Diffraction patterns can also be observed in other scenarios, such as when light passes through a single slit or when waves pass through a narrow opening.
  • The resulting pattern is characterized by a central maximum and a series of alternating dark and light fringes.
  • The intensity of the fringes decreases as the distance from the central maximum increases.

Slide 22

  • Single Slit Diffraction Equation:
    • The intensity of the diffraction pattern of light passing through a single slit can be calculated using the equation:
      • I = I₀ * [sin(θ) / θ]²
    • In this equation:
      • I is the intensity,
      • I₀ is the intensity at the center (θ = 0),
      • θ is the angle between the direction of the diffracted light and the direction of the incident light.

Slide 23

  • Example: Single Slit Diffraction Pattern
    • A light source with a wavelength of 600 nm is directed towards a single slit with a width of 0.1 mm. Calculate the angular position of the first minimum in the diffraction pattern.
  • Solution:
    • Given: λ = 600 nm, d = 0.1 mm.
    • Using the formula for the first minimum in the single slit diffraction pattern: sin(θ) = 1.22 * λ / d
    • Substituting the given values: sin(θ) = 1.22 * 600 nm / 0.1 mm
    • Converting nm to meters and mm to meters: sin(θ) = 1.22 * 600 × 10⁻⁹ m / 0.1 × 10⁻³ m
    • Simplifying the expression: sin(θ) = 7.32 * 10⁻⁶
    • Taking the inverse sine: θ ≈ sin⁻¹(7.32 * 10⁻⁶)

Slide 24

  • Example (continued):
    • Evaluating the value using a scientific calculator: θ ≈ 0.42°
  • Therefore, the angular position of the first minimum in the diffraction pattern is approximately 0.42°.

Slide 25

  • Diffraction of Sound Waves:
    • Diffraction is not limited to light waves; it can also occur with other types of waves, such as sound waves.
    • When sound waves encounter an obstacle or pass through an opening, they can diffract and spread out.
    • The magnitude of diffraction depends on the wavelength of the sound waves and the size of the obstacle or opening.

Slide 26

  • Example: Diffraction of Sound Waves
    • A sound wave with a wavelength of 0.5 meters passes through a narrow opening that is 1 meter wide. Calculate the angle at which the first diffracted maximum occurs.
  • Solution:
    • Given: λ = 0.5 m, d = 1 m.
    • Using the formula for the first maximum in the diffraction pattern: sin(θ) = λ / d
    • Substituting the given values: sin(θ) = 0.5 m / 1 m
    • Simplifying the expression: sin(θ) = 0.5
    • Taking the inverse sine: θ = sin⁻¹(0.5)

Slide 27

  • Example (continued):
    • Evaluating the value using a scientific calculator: θ ≈ 30°
  • Therefore, the angle at which the first diffracted maximum occurs is approximately 30°.

Slide 28

  • Diffraction and Polarization:
    • Diffraction can also affect the polarization of light waves.
    • When unpolarized light passes through a narrow opening or encounters an obstacle, it becomes partially polarized.
    • The degree of polarization depends on the size of the opening or obstacle and the wavelength of light.

Slide 29

  • Example: Diffraction and Polarization
    • Unpolarized light with a wavelength of 500 nm passes through a narrow slit that is 0.1 mm wide. Calculate the degree of polarization of the diffracted light.
  • Solution:
    • Given: λ = 500 nm, d = 0.1 mm.
    • Using the formula for the degree of polarization: P = 2 sin²(θ) / (1 + cos²(θ))
    • Using the formula for the first maximum in the single slit diffraction pattern: sin(θ) = λ / d
    • Substituting the given values: sin(θ) = 500 nm / 0.1 mm
    • Evaluating the value: sin(θ) = 0.05
    • Substituting sin(θ) in the degree of polarization formula: P = 2 (0.05)² / (1 + cos²(θ))

Slide 30

  • Example (continued):
    • Evaluating the value using a scientific calculator with θ = sin⁻¹(0.05): P ≈ 0.08
  • Therefore, the degree of polarization of the diffracted light is approximately 0.08.
  • These are some of the key concepts related to diffraction in the context of intensity distribution on the focal plane. I hope this lecture has provided you with a good understanding of the topic.