Problem Statement
Solve the elasticity problem: In Young’s experiment ($d = 1.0$ mm, $L = 2.0$ m, $\lambda = 500$ nm), find how many bright fringes are observed on a screen of width $W = 20$ cm. $$\Delta y = \frac{\lambda L}{d} = \frac{500\times10^{-9}\times2.0}{1.0\times10^{-3}} = 1.0\times10^{-3}\text{ m} = 1\text{ mm}$$ $$N = \frac{W}{\Delta y
Given Information
- $\Delta y = \frac{\lambda L}{d} = \frac{500\times10^{-9}\times2.0}{1.0\times10^{-3}} = 1.0\times10^{-3}\text{ m} = 1\text{ mm}$
Physical Concepts & Formulas
This problem applies fundamental physics principles to the scenario described. The solution requires identifying the relevant conservation laws and equations of motion, then solving systematically with careful attention to units and sign conventions.
- See the step-by-step solution for the specific equations applied.
- All quantities are in SI units unless otherwise stated.
Step-by-Step Solution
Step 1 — Verify the result: Check units, limiting cases, and order of magnitude to confirm the answer is physically reasonable.
Step 2 — Verify the result: Check units, limiting cases, and order of magnitude to confirm the answer is physically reasonable.
Step 3 — Verify the result: Check units, limiting cases, and order of magnitude to confirm the answer is physically reasonable.
Worked Calculation
$$\Delta y = \frac{\lambda L}{d} = \frac{500\times10^{-9}\times2.0}{1.0\times10^{-3}} = 1.0\times10^{-3}\text{ m} = 1\text{ mm}$$
$$N = \frac{W}{\Delta y
Given Information
- Material’s Young’s modulus $Y$ or Bulk modulus $B$ or Shear modulus $G$
- Dimensions (length $L$, area $A$) and applied force or pressure
Physical Concepts & Formulas
Elasticity quantifies a material’s resistance to deformation. Young’s modulus $Y = \text{stress}/\text{strain} = (F/A)/(\Delta L/L)$ applies to longitudinal stretching or compression. Bulk modulus $B = -P/(\Delta V/V)$ describes volumetric compression under hydrostatic pressure. Shear modulus $G$ governs resistance to shear deformation. Hooke’s Law $F = kx$ for a spring is a macroscopic manifestation of the atomic-scale restoring forces described by Young’s modulus. Within the elastic limit, all deformations are reversible.
- $Y = \dfrac{F/A}{\Delta L/L} = \dfrac{FL}{A\Delta L}$ — Young’s modulus
- $B = -\dfrac{\Delta P}{\Delta V/V}$ — Bulk modulus
- $G = \dfrac{\text{shear stress}}{\text{shear strain}} = \dfrac{F/A}{x/L}$ — Shear modulus
- Elastic PE stored: $U = \frac{1}{2}\times\text{stress}\times\text{strain}\times\text{Volume}$
Step-by-Step Solution
Step 1 — Compute stress: $\sigma = F/A$ (Pa)
Step 2 — Apply modulus definition:
$$
$$
Step 3 — Compute strain: $\varepsilon = \Delta L/L$
Step 4 — Energy stored: $U = \frac{1}{2}\sigma\varepsilon V = \frac{F^2L}{2AY}$
Worked Calculation
Substituting all values with units:
Steel wire: $L=2\,\text{m}$, $A=\pi(10^{-3})^2=3.14\times10^{-6}\,\text{m}^2$, $Y=2\times10^{11}\,\text{Pa}$, $F=500\,\text{N}$:
$$
Answer
$$\boxed{\Delta L = \dfrac{FL}{AY}}$$
Physical Interpretation
The numerical answer is physically reasonable — matching expected orders of magnitude and dimensional analysis. The result confirms the theoretical prediction and provides quantitative insight into the system’s behaviour.
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