Exercises — Young's double slit — fringe width derivation

KYA chahiye: $\beta$, consecutive bright fringes ki spacing. YEH formula kyun: humein seedha $\lambda ,D,d$ pata hai, toh $\beta =\frac{\lambda D}{d}$ use karo. Pehle SI mein convert karo ($1\text{nm}=10^{-9}\text{m}$, $1\text{mm}=10^{-3}\text{m}$) taaki answer metres mein aaye: $\beta =\frac{(500\times 10^{-9})(1.0)}{0.25\times 10^{-3}}=\frac{5\times 10^{-7}}{2.5\times 10^{-4}}=2\times 10^{-3}\text{m}=2\text{mm}$ Answer: $\beta =2\text{mm}$.

KYA: $P$ ko classify karo. KYU: bright/dark ka rule purely $\Delta$ ke terms mein stated hai.

• Bright ke liye $\Delta =n\lambda$ chahiye (poora number of wavelengths).
• Dark ke liye $\Delta =(n+\frac{1}{2})\lambda$ chahiye (half-integer number). Yahan $\Delta =1.5\lambda =(1+\frac{1}{2})\lambda$, ek half-integer multiple hai ⇒ dark fringe. Answer: dark.

$y_n$ kyun, $\beta$ nahi: humein ek specific fringe ki position batayi gayi hai ($n=4$), toh $y_n=\frac{n\lambda D}{d}$ use karo, phir $\lambda$ ke liye solve karo: $\lambda =\frac{y_{n}d}{nD}=\frac{(5.6\times 10^{-3})(0.3\times 10^{-3})}{4\times 0.6}=\frac{1.68\times 10^{-6}}{2.4}=7\times 10^{-7}\text{m}$ Answer: $\lambda =700\text{nm}$ (red light).

KYA: centre ke dono taraf $4.5\text{mm}$ half-width wali window mein bright bands ki sankhya. Bright fringes $y_n=n\beta$ par hoti hain. Humein $|n\beta |\le 4.5\text{mm}$ chahiye: $|n|\le \frac{4.5}{1.2}=3.75\Rightarrow n=-3,-2,-1,0,1,2,3$ Yeh $7$ values hain. Answer: $7$ bright fringes.

Ratios kyun: $\beta =\frac{\lambda D}{d}$, toh $\beta \propto \frac{D}{d}$ fixed $\lambda$ par. Raw numbers ki jagah multiplying factors track karo: $\frac{{\beta }_1}{{\beta }_0}=\frac{D_1/d_1}{D_0/d_0}=\frac{(3D_0)/(d_0/2)}{D_0/d_0}=3\times 2=6$ ${\beta }_1=6\times 1.5\text{mm}=9\text{mm}$ Answer: ${\beta }_1=9\text{mm}$ — cheh guna zyada wide.

KYA: ek aisa point jahan ${\lambda }_1$ ki $n_1$-th bright fringe ${\lambda }_2$ ki $n_2$-th bright fringe par land kare. $y_n$ equate kyun: overlap matlab positions match karein: $\frac{n_1{\lambda }_{1}D}{d}=\frac{n_2{\lambda }_{2}D}{d}\Rightarrow n_1{\lambda }_1=n_2{\lambda }_2\Rightarrow \frac{n_1}{n_2}=\frac{{\lambda }_2}{{\lambda }_1}=\frac{600}{480}=\frac{5}{4}$ Smallest whole numbers: $n_1=5,n_2=4$. Coincidence height: $y=\frac{n_1{\lambda }_{1}D}{d}=\frac{5(480\times 10^{-9})(1.0)}{0.5\times 10^{-3}}=\frac{2.4\times 10^{-6}}{5\times 10^{-4}}=4.8\times 10^{-3}\text{m}$ Answer: $y=4.8\text{mm}$ (jo blue ki 5th fringe = orange ki 4th fringe hai).

$\lambda$ kyun shrink hoti hai: frequency source se fixed hoti hai; medium mein wave $n$ factor se slow ho jaati hai, toh uski wavelength shrink ho jaati hai: ${\lambda }^{\prime }=\lambda /n$. (Dekho Refractive index.) Kyunki $\beta \propto \lambda$ aur $D,d$ unchanged hain: ${\beta }^{\prime }=\frac{{\lambda }^{\prime }D}{d}=\frac{{\beta }_{\text{air}}}{n}=\frac{1.8}{1.5}=1.2\text{mm}$ Physical picture: chota ruler matlab path-difference milestones $\lambda ,2\lambda ,\dots$ jaldi aa jaate hain, toh bright bands paas-paas aa jaate hain. Answer: ${\beta }^{\prime }=1.2\text{mm}$.

Sheet kya karti hai: yeh $S_1$ path ko $(n-1)t$ se optically longer banati hai. Central fringe wahan hoti hai jahan total path difference zero ho, toh geometric path ab usi amount se $S_1$ ko favour karna chahiye. Geometric path difference ko extra optical length ke barabar set karo: $\frac{y_{0}d}{D}=(n-1)t$ KYU: ${\Delta }_{\text{geometry}}=\frac{y_{0}d}{D}$ ko added $(n-1)t$ cancel karna chahiye. $y_0=\frac{(n-1)tD}{d}=\frac{(0.5)(10\times 10^{-6})(1.2)}{0.6\times 10^{-3}}=\frac{6\times 10^{-6}}{6\times 10^{-4}}=1.0\times 10^{-2}\text{m}$ Answer: $y_0=10\text{mm}$, covered slit $S_1$ ki side ki taraf shift hua.

Ratios kyun phir: $d,D$ fixed hone par $\beta \propto \lambda$, toh $\frac{{\lambda }_X}{{\lambda }_R}=\frac{{\beta }_X}{{\beta }_R}=\frac{1.5}{2.1}=\frac{5}{7}$ ${\lambda }_X=\frac{5}{7}\times 700=500\text{nm}$ Answer: ${\lambda }_X=500\text{nm}$ (green).

Step 1 — har side par kitni fringes. $9$ bright fringes jo centre ke around symmetric hain, matlab ek central fringe ($n=0$) plus dono side par $4$ ($n=\pm 1,\dots ,\pm 4$). Toh outermost $n=4$ hai $y_4=6\text{mm}$ par. Step 2 — $\beta$ se relate karo. Kyunki $y_4=4\beta$: $\beta =\frac{y_4}{4}=\frac{6\text{mm}}{4}=1.5\text{mm}$ Step 3 — $d$ ke liye solve karo $\beta =\frac{\lambda D}{d}$ se: $d=\frac{\lambda D}{\beta }=\frac{(546\times 10^{-9})(1.5)}{1.5\times 10^{-3}}=\frac{8.19\times 10^{-7}}{1.5\times 10^{-3}}=5.46\times 10^{-4}\text{m}$ Answer: $d\approx 0.546\text{mm}$.