Exercises — Speed of light c = 1 - √(ε₀ μ₀)

$c=\frac{1}{\sqrt{{\epsilon }_0{\mu }_0}}$

• ${\epsilon }_0$ = free space ki permittivity — khali jagah kitni aasani se electric field ko permit (support) karti hai.
• ${\mu }_0$ = free space ki permeability — khali jagah kitni aasani se magnetic field ko support karti hai. Dono vacuum ki "stiffnesses" hain. Zyada stiffness → slower relay → chhota $c$, isliye ye dono bilkul theek denominator mein hain.

Product ${\epsilon }_0{\mu }_0$ ki units hain ${\text{s}}^2/{\text{m}}^2$ (yani $(\text{time}/\text{distance}{)}^2$, matlab ek slowness squared).

• $\sqrt{{\epsilon }_0{\mu }_0}$ → units $\text{s}/\text{m}$ = ek slowness, speed nahi. ✗
• $1/\sqrt{{\epsilon }_0{\mu }_0}$ → units $\text{m}/\text{s}$ = ek speed. ✓ Toh reciprocal form hi ek aisi cheez hai jo speed ho sakti hai.

Step 1 — constants ko multiply karo. KYU: formula ko pehle product ${\epsilon }_0{\mu }_0$ chahiye. ${\mu }_0=4\pi \times 10^{-7}=1.2566\times 10^{-6}{\text{T m A}}^{-1}$ ${\epsilon }_0{\mu }_0=(8.85\times 10^{-12})(1.2566\times 10^{-6})=1.112\times 10^{-17}{\text{s}}^2{\text{m}}^{-2}$ Step 2 — square root lo. KYU: formula mein denominator mein  ​ hai. $\sqrt{1.112\times 10^{-17}}=3.335\times 10^{-9}{\text{s m}}^{-1}$ Step 3 — invert karo. KYU: $c$ us slowness ka reciprocal hai. $c=\frac{1}{3.335\times 10^{-9}}=2.998\times 10^8{\text{m s}}^{-1}\approx 3.00\times 10^8{\text{m s}}^{-1}$

Step 1 — index. KYU: $n=\sqrt{{\epsilon }_r{\mu }_r}$ wohi derivation run karke aati hai material constants ke saath (dekho Refractive Index and n = c/v). $n=\sqrt{2.25\times 1}=1.5$ Step 2 — speed. KYU: $v=c/n$. $v=\frac{3.00\times 10^8}{1.5}=2.00\times 10^8{\text{m s}}^{-1}$

KYU ye tool: ek plane EM wave ke liye wohi Maxwell equations $E=cB$ ko har instant force karti hain (dekho Electromagnetic Waves). $E_0=c{B}_0=(3.00\times 10^8)(1.0\times 10^{-8})=3.0{\text{V m}}^{-1}$ Electric amplitude numerically magnetic wale se $3\times 10^8$ times zyada hota hai — wo bada factor bas $c$ hai.

Galti ye hai ki pre-Maxwell Ampère law $\nabla \times \vec{B}={\mu }_0\vec{J}$ use ki ja rahi hai. Maxwell ne displacement current term add kiya tha: $\nabla \times \vec{B}={\mu }_0\vec{J}+{\mu }_0{\epsilon }_0\frac{\partial \vec{E}}{\partial t}.$ Vacuum mein $\vec{J}=0$ hai, lekin doosra term tab bhi survive karta hai jab bhi $\vec{E}$ time mein change karta hai. Yahi term exactly wave-equation derivation ke Step 3 mein aata hai; iske bina equation ${\nabla }^2\vec{E}={\mu }_0{\epsilon }_0{\partial }^2\vec{E}/\partial t^2$ kabhi nahi banti aur light ka koi sawaal hi nahi. Dekho Displacement Current aur Ampère–Maxwell Law.

Standard 1-D Wave Equation hai: $\frac{{\partial }^{2}f}{\partial x^2}=\frac{1}{v^2}\frac{{\partial }^{2}f}{\partial t^2}.$ ${\partial }^{2}E/\partial x^2={\mu }_0{\epsilon }_0{\partial }^{2}E/\partial t^2$ se term-by-term match karke: $\frac{1}{v^2}={\mu }_0{\epsilon }_0\Rightarrow v=\frac{1}{\sqrt{{\mu }_0{\epsilon }_0}}=c.$ Units: ${\epsilon }_0{\mu }_0$ mein ${\text{s}}^2{\text{m}}^{-2}$ hoti hai, toh $1/\sqrt{{\epsilon }_0{\mu }_0}$ mein $\sqrt{{\text{m}}^2{\text{s}}^{-2}}={\text{m s}}^{-1}$ hoti hai. ✓ Neeche diya picture ${\partial }^2/\partial x^2$ aur ${\partial }^2/\partial t^2$ ke do curves dikhata hai jo proportion mein force hote hain.

Pehla material: $n=\sqrt{4\cdot 1}=2$, toh $v=c/2$ — factor 2 se slow hoti hai. Doosra material: $n=\sqrt{1\cdot 4}=2$, toh bhi $v=c/2$ — same slowdown. Revelation: ${\epsilon }_r$ aur ${\mu }_r$ dono square root ke neeche unke product ke through symmetrically enter karte hain. Sirf product ${\epsilon }_r{\mu }_r$ hi speed control karta hai; ek given slowdown ya toh extra electric "give" se ya extra magnetic "give" se equally produce ho sakta hai.

${\epsilon }_0$ Coulomb's law se fix hota hai (charges ke beech force); ${\mu }_0$ current-carrying wires ke beech force se fix hota hai. Kisi bhi experiment mein light involve nahi hai. Combine karne par: $c=\frac{1}{\sqrt{(8.85\times 10^{-12})(4\pi \times 10^{-7})}}=3.00\times 10^8{\text{m s}}^{-1}.$ Yeh optically measured speed of light se match karta hai. Kyunki dono constants purely electricity aur magnetism se aaye, yeh match ek proof hai ki light ek electromagnetic phenomenon hai — teen pehle alag subjects ka ek unification.

(a) $E=cB$ se: $B_0=E_0/c=(1.0\times 10^3)/(3.00\times 10^8)=3.33\times 10^{-6}\text{T}$. (b) $E_0/B_0$ ki units hain $\frac{\text{V/m}}{\text{T}}$. Kyunki $1\text{T}=1{\text{V⋅s⋅m}}^{-2}$, $\frac{\text{V/m}}{{\text{V⋅s⋅m}}^{-2}}=\frac{\text{V}}{\text{m}}\cdot \frac{{\text{m}}^2}{\text{V⋅s}}=\frac{\text{m}}{\text{s}}.$ ✓ Toh $E_0/B_0=c=3.00\times 10^8{\text{m s}}^{-1}$, (a) se consistent hai.

KYU: $c\propto 1/\sqrt{{\epsilon }_0{\mu }_0}$, toh ${\epsilon }_0$ ko 10 se multiply karne par $c$ mein $1/\sqrt{10}$ ka multiplication hota hai. $c^{\prime }=\frac{c}{\sqrt{10}}=0.316c\approx 9.49\times 10^7{\text{m s}}^{-1}.$ Ek "stiffer" (zyada permitting) vacuum mein field-relay sirf tehai speed se aage badh sakta hai.

Source-free Maxwell equations (dekho Maxwell's Equations): $\nabla \cdot \vec{E}=0,\nabla \cdot \vec{B}=0,\nabla \times \vec{E}=-\frac{\partial \vec{B}}{\partial t},\nabla \times \vec{B}={\mu }_0{\epsilon }_0\frac{\partial \vec{E}}{\partial t}.$ Step 1 — Ampère–Maxwell law ka curl lo. WHAT: $\nabla \times \vec{B}={\mu }_0{\epsilon }_0\partial \vec{E}/\partial t$ par $\nabla \times$ apply karo. KYU: baad mein $\nabla \times \vec{E}$ ki doosri equation substitute karke $\vec{B}$ mein sirf ek equation isolate karna hai. $\nabla \times (\nabla \times \vec{B})={\mu }_0{\epsilon }_0\frac{\partial }{\partial t}(\nabla \times \vec{E}).$ Step 2 — vector identity. WHAT: $\nabla \times (\nabla \times \vec{B})=\nabla (\nabla \cdot \vec{B})-{\nabla }^2\vec{B}$. KYU: double curl ko Laplacian mein convert karta hai. Kyunki $\nabla \cdot \vec{B}=0$ hai, pehla term khatam ho jaata hai: $-{\nabla }^2\vec{B}={\mu }_0{\epsilon }_0\frac{\partial }{\partial t}(\nabla \times \vec{E}).$ Step 3 — Faraday substitute karo $\nabla \times \vec{E}=-\partial \vec{B}/\partial t$. KYU: equations ko couple karta hai aur time derivative laata hai: $-{\nabla }^2\vec{B}={\mu }_0{\epsilon }_0\frac{\partial }{\partial t}\left(-\frac{\partial \vec{B}}{\partial t}\right)=-{\mu }_0{\epsilon }_0\frac{{\partial }^2\vec{B}}{\partial t^2}.$ Step 4 — clean up: ${\nabla }^2\vec{B}={\mu }_0{\epsilon }_0\frac{{\partial }^2\vec{B}}{\partial t^2}.$ Form mein $\vec{E}$ equation jaisa identical hai ⇒ same speed $v=1/\sqrt{{\mu }_0{\epsilon }_0}=c$. Toh $\vec{E}$ aur $\vec{B}$ ek hi speed par saath locked hokar travel karte hain. Figure relay dikhata hai: har field ka change doosre ko source karta hai.

Step 1 — differentiate karo. KYU: 1-D mein Faraday $E$ ke space slope ko $B$ ke time rate se relate karta hai. $\frac{\partial E}{\partial x}=E_{0}k\cos (kx-\omega t),-\frac{\partial B}{\partial t}=+B_0\omega \cos (kx-\omega t).$ Step 2 — common $\cos (kx-\omega t)$ ke coefficients equate karo: $E_{0}k=B_0\omega \Rightarrow \frac{E_0}{B_0}=\frac{\omega }{k}.$ Step 3 — phase speed identify karo. KYU: $\sin (kx-\omega t)$ ke liye wave speed $\omega /k=c$ hoti hai. Isliye $\frac{E_0}{B_0}=c\Rightarrow E_0=c{B}_0.■$

(a) $E_0=c{B}_0=(3.00\times 10^8)(2.0\times 10^{-6})=6.0\times 10^2{\text{V m}}^{-1}=600{\text{V m}}^{-1}$. (b) $n=\sqrt{2.25}=1.5$, toh $v=c/n=(3.00\times 10^8)/1.5=2.00\times 10^8{\text{m s}}^{-1}$. (c) Frequency $f$ boundary par fixed rehti hai; $v=f\lambda$, toh $\lambda \propto v$: $\frac{{\lambda }_{\text{medium}}}{{\lambda }_{\text{vacuum}}}=\frac{v}{c}=\frac{1}{n}=\frac{1}{1.5}=\mathrm{0.667.}$ Wavelength same factor se shrink hoti hai jis factor se speed drop hoti hai.

• L2.1: $c\approx 3.00\times 10^8{\text{m s}}^{-1}$
• L2.2: $n=1.5$, $v=2.00\times 10^8{\text{m s}}^{-1}$
• L2.3: $E_0=3.0{\text{V m}}^{-1}$
• L3.3: dono factor 2 se slow ($n=2$)
• L4.2: $B_0=3.33\times 10^{-6}\text{T}$
• L4.3: $c^{\prime }=0.316c\approx 9.49\times 10^7{\text{m s}}^{-1}$
• L5.3: $E_0=600{\text{V m}}^{-1}$; $v=2.00\times 10^8{\text{m s}}^{-1}$; $\lambda$ ratio $=0.667$