Problem Statement
Solve the elasticity problem: In Young’s double slit experiment: $d=1.0$ mm, $D=1.0$ m, $\lambda=589$ nm. Find fringe width. $\beta=\lambda D/d$ Step 1: $\beta=\lambda D/d=589\times10^{-9}\times1.0/10^{-3}=5.89\times10^{-4}$ m $=0.589$ mm. $$\boxed{\beta=0.589\text{ mm}}$$
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:
$$\Delta L = \frac{FL}{AY}$$
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: $L=2\,\text{m}$, $A=\pi(10^{-3})^2=3.14\times10^{-6}\,\text{m}^2$, $Y=2\times10^{11}\,\text{Pa}$, $F=500\,\text{N}$:
$$\Delta L = \frac{FL}{AY} = \frac{500\times2}{3.14\times10^{-6}\times2\times10^{11}} = \frac{1000}{6.28\times10^5} = 1.59\times10^{-3}\,\text{m} \approx 1.6\,\text{mm}$$
Answer
$$\boxed{\Delta L = \dfrac{FL}{AY}}$$
Physical Interpretation
A 1 mm diameter steel wire stretches only 1.6 mm under 500 N (about 50 kg weight) over 2 m — steel’s enormous Young’s modulus ($2\times10^{11}\,\text{Pa}$) makes it very stiff. Rubber has $Y \sim 10^6\,\text{Pa}$, 200,000 times softer. This is why steel cables support bridges while rubber bands stretch easily. Beyond the elastic limit, the wire would undergo plastic deformation and not return to its original length.
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