Young’s Double Slit Experiment

Evidence for the Wave-like Nature of Light

• Young’s double-slit experiment demonstrates the wave-like nature of light through the phenomenon of interference.
• When light passes through two narrow, closely spaced slits, it creates an interference pattern on a screen behind the slits.
• This pattern consists of alternating bright fringes (constructive interference) and dark fringes (destructive interference).
• The formation of this interference pattern is only possible if light behaves as a wave, capable of superposition and interference.
• This experiment challenged the particle theory of light and supported the wave theory proposed by Huygens and Hooke.

KEY TAKEAWAY: The observation of an interference pattern in Young’s double-slit experiment is direct evidence that light exhibits wave-like properties.

Constructive and Destructive Interference

• Coherent waves are required for a stable interference pattern. Coherent waves have the same frequency, wavelength, and a constant phase relationship.
• Constructive interference occurs when the path difference between two coherent waves is an integer multiple of the wavelength ($n\lambda$), where $n = 0, 1, 2, …$. This results in a bright fringe.
  • Path difference = $n\lambda$
• Destructive interference occurs when the path difference between two coherent waves is a half-integer multiple of the wavelength $((n + \frac{1}{2})\lambda)$, where $n = 0, 1, 2, …$. This results in a dark fringe.
  • Path difference = $(n + \frac{1}{2})\lambda$

REMEMBER: Bright fringes result from constructive interference (path difference = $n\lambda$), while dark fringes result from destructive interference (path difference = $(n + \frac{1}{2})\lambda$).

Effect of Wavelength, Distance of Screen, and Slit Separation on Interference Patterns

• The fringe spacing (distance between adjacent bright or dark fringes) in Young’s double-slit experiment is given by the formula:

  $$y = \frac{m\lambda L}{d}$$

  Where:
  * $y$ = distance from the central maximum to the $m^{th}$ fringe
  * $m$ = fringe order (0 for central maximum, 1 for first bright fringe, etc.)
  * $\lambda$ = wavelength of light
  * $L$ = distance from the slits to the screen
  * $d$ = separation between the two slits

  Or, the fringe spacing, $\Delta x$, (the distance between two adjacent bright fringes) can be written as:
  $$\Delta x = \frac{\lambda L}{d}$$

• Effect of Wavelength ($\lambda$):

  • As the wavelength increases, the fringe spacing increases. Longer wavelengths (e.g., red light) produce wider fringe patterns than shorter wavelengths (e.g., blue light).
• Effect of Distance of Screen ($L$):
  • As the distance between the slits and the screen increases, the fringe spacing increases. The fringes become more spread out.
• Effect of Slit Separation ($d$):
  • As the slit separation increases, the fringe spacing decreases. The fringes become more closely spaced.
• Condition for Validity ($L >> d$):
  • The formula $y = \frac{m\lambda L}{d}$ is valid when the distance from the slits to the screen ($L$) is much greater than the slit separation ($d$). This ensures that the angle to the fringes is small, allowing for the approximation.

EXAM TIP: Be able to rearrange the formula $\Delta x = \frac{\lambda L}{d}$ to solve for any of the variables ($\lambda$, $L$, $d$, or $\Delta x$) given the other values. Pay attention to units!

Summary Table

COMMON MISTAKE: Forgetting to convert all measurements to consistent units (e.g., meters) before applying the formula. Often, slit separation is given in millimeters or micrometers, while wavelength is given in nanometers.

Worked Example

Light of wavelength 600 nm is incident on two slits that are 0.1 mm apart. A screen is placed 2.0 m from the slits.

a) Calculate the fringe spacing.

$\lambda = 600 \text{ nm} = 600 \times 10^{-9} \text{ m}$
$d = 0.1 \text{ mm} = 0.1 \times 10^{-3} \text{ m}$
$L = 2.0 \text{ m}$

$$\Delta x = \frac{\lambda L}{d} = \frac{(600 \times 10^{-9} \text{ m})(2.0 \text{ m})}{0.1 \times 10^{-3} \text{ m}} = 0.012 \text{ m} = 12 \text{ mm}$$

b) What is the distance from the central maximum to the third bright fringe?

$m = 3$
$$y = \frac{m\lambda L}{d} = \frac{3(600 \times 10^{-9} \text{ m})(2.0 \text{ m})}{0.1 \times 10^{-3} \text{ m}} = 0.036 \text{ m} = 36 \text{ mm}$$

STUDY HINT: Draw diagrams to visualize the experiment and the relationships between the variables. Label the diagram with the given values to help with problem-solving.

Key Concepts Recap

• Young’s double-slit experiment provides evidence for the wave nature of light through the observation of interference patterns.
• Constructive interference occurs when the path difference is $n\lambda$, resulting in bright fringes.
• Destructive interference occurs when the path difference is $(n + \frac{1}{2})\lambda$, resulting in dark fringes.
• The fringe spacing depends on the wavelength of light, the distance to the screen, and the slit separation, according to the formula $\Delta x = \frac{\lambda L}{d}$.

VCAA FOCUS: VCAA exam questions often combine conceptual understanding of interference with quantitative problem-solving using the fringe spacing formula. Be prepared to explain the experiment and perform calculations.