Physics Study Notes: Diffraction, Thermodynamics, and pV Diagrams

Single-Slit Diffraction

Dark Fringes in Single-Slit Diffraction

When monochromatic light passes through a single narrow slit, it produces a diffraction pattern of bright and dark fringes on a distant screen. The positions of the dark fringes (minima) are determined by the condition for destructive interference.

• Condition for Dark Fringes: The minima occur at angles θ where the path difference causes destructive interference, given by:

$a \sin\theta = m\lambda$

• Here, a is the slit width, λ is the wavelength of light, and m is a nonzero integer (±1, ±2, ...).

• Maximum Number of Dark Fringes: The largest possible value for $|\sin\theta|$ is 1. Thus, the maximum order m is:

$m_{\text{max}} = \frac{a}{\lambda}$

• For both sides of the central maximum, the total number of dark fringes is $2m_{\text{max}}$.

• Example Calculation: For $a = 0.0666$ mm $= 6.66 \times 10^{-5}$ m and $\lambda = 585$ nm $= 5.85 \times 10^{-7}$ m:

$m_{\text{max}} = \frac{6.66 \times 10^{-5} \text{ m}}{5.85 \times 10^{-7} \text{ m}} \approx 113.8$

• Thus, $m$ runs from $-113$ to $+113$ (excluding $m=0$), giving $226$ dark fringes.

• Angle of Most Distant Dark Fringe: For $m = \pm 113$:

$\sin\theta = \frac{113 \times 5.85 \times 10^{-7}}{6.66 \times 10^{-5}} \approx 0.99296$

$\theta = \arcsin(0.99296) \approx 83.0^\circ$

• Physical Interpretation: As the slit width decreases, fewer dark fringes are observed. If the slit is narrower than the wavelength, no dark fringes appear and the central maximum fills the screen.

Thermodynamics of an Ideal Gas: pV Diagrams

pV Diagram and Thermodynamic Processes

A pV diagram plots the pressure (p) of a gas against its volume (V). Different paths on this diagram represent different thermodynamic processes. For a monatomic ideal gas, the internal energy, heat, and work can be analyzed for each segment.

• Key Definitions:

  • Isobaric Process: Pressure remains constant.

  • Isochoric Process: Volume remains constant.

  • Isothermal Process: Temperature remains constant.

• Ideal Gas Law:

$pV = nRT$

• Where n is the number of moles, R is the gas constant ($8.315$ J/mol·K), T is temperature.

• Internal Energy Change: For a monatomic ideal gas:

$\Delta U = nC_v\Delta T$

• Where $C_v = \frac{3}{2}R$ for monatomic gases.

• Isothermal Process: $\Delta U = 0$, so $Q = W = nRT \ln\left(\frac{V_f}{V_i}\right)$

Example: Analysis of a pV Diagram

Given: 3.25 moles of helium gas undergoes a cycle with segments ab (isobaric), bc (isochoric), and ca (isothermal).

• Pressures and Temperatures:

  • At point a: $p_a = 8.0 \times 10^5$ Pa, $V_a = 0.010$ m³

  • At point b: $p_b = 8.0 \times 10^5$ Pa, $V_b = 0.040$ m³

  • At point c: $p_c = 2.0 \times 10^5$ Pa, $V_c = 0.040$ m³

• Temperatures:

$T_a = \frac{p_a V_a}{nR} = \frac{8.0 \times 10^5 \times 0.010}{3.25 \times 8.315} = 296 \text{ K}$ $T_b = \frac{p_b V_b}{nR} = \frac{8.0 \times 10^5 \times 0.040}{3.25 \times 8.315} = 1184 \text{ K}$ $T_c = \frac{p_c V_c}{nR} = \frac{2.0 \times 10^5 \times 0.040}{3.25 \times 8.315} = 296 \text{ K}$

• Heat Transfer (Q) and Internal Energy Change (ΔU):

  • ab (isobaric): $Q = nC_p\Delta T$, $\Delta U = nC_v\Delta T$

  • bc (isochoric): $Q = nC_v\Delta T$, $\Delta U = nC_v\Delta T$

  • ca (isothermal): $Q = nRT \ln\left(\frac{V_a}{V_c}\right)$, $\Delta U = 0$

$Q_{ab} = (3.25)(8.315)(1184 - 296) = 6.00 \times 10^4 \text{ J}$ $Q_{bc} = (3.25)(8.315)(296 - 1184) = -3.60 \times 10^4 \text{ J}$ $Q_{ca} = (3.25)(8.315)(296) \ln\left(\frac{0.010}{0.040}\right) = -1.11 \times 10^4 \text{ J}$

• Internal Energy Changes:

$\Delta U_{ab} = (3.25)(\frac{3}{2} \times 8.315)(1184 - 296) = 3.60 \times 10^4 \text{ J}$ $\Delta U_{bc} = (3.25)(\frac{3}{2} \times 8.315)(296 - 1184) = -3.60 \times 10^4 \text{ J}$ $\Delta U_{ca} = 0$

• Direction of Heat Flow:

  • ab: Heat enters the gas.

  • bc: Heat leaves the gas.

  • ca: Heat leaves the gas.

• Energy Changes:

  • ab: Internal energy increases.

  • bc: Internal energy decreases.

  • ca: No change in internal energy.

Summary Table: Thermodynamic Quantities for Each Segment

Segment	Process	ΔU (J)	Q (J)	Heat Flow	ΔU Direction
ab	Isobaric	+3.60 × 10⁴	+6.00 × 10⁴	Enters	Increases
bc	Isochoric	−3.60 × 10⁴	−3.60 × 10⁴	Leaves	Decreases
ca	Isothermal	0	−1.11 × 10⁴	Leaves	No change

Additional info: For any closed path on a pV diagram, the net change in internal energy is zero, since the system returns to its initial state.