Exercises — Diffraction grating — condition for maxima

WHAT: "Lines per length" ko spacing mein convert karo. WHY: Grating equation ko $d$ (ek distance) chahiye, raw line count nahi. $d=\frac{1}{n}=\frac{1}{300{\text{mm}}^{-1}}=\frac{1}{300}\text{mm}=3.33\times 10^{-3}\text{mm}$ Metres mein convert karo ($1\text{mm}=10^{-3}\text{m}$): $d=3.33\times 10^{-3}\times 10^{-3}\text{m}=3.33\times 10^{-6}\text{m}$ Answer: $d\approx 3.33\times 10^{-6}\text{m}$ (yaani $3333\text{nm}$).

WHAT: Ek ray jo extra distance travel karti hai, woh state karo. WHY: Yahi poori grating equation ki neev hai. Dono rays parallel hain, isliye unka fark sirf woh chhota sa segment hai jo ek slit se doosri ray par perpendicular daalte waqt cut hota hai. Woh segment ek right triangle ki opposite side hai jiska hypotenuse $d$ hai aur angle $\theta$ hai: $\Delta =d\sin \theta$ Answer: $\Delta =d\sin \theta$, right triangle se jiska hypotenuse $d$ hai (upar figure mein red segment dekho).

Step 1 — order choose karo. First order matlab $m=1$. Step 2 — unknown angle ke liye rearrange karo. $\sin {\theta }_1=\frac{m\lambda }{d}=\frac{1\times 550\times 10^{-9}}{2.0\times 10^{-6}}=0.275$ Kyun rearrange karte hain? $\theta$ woh cheez hai jo chahiye, isliye pehle $\sin \theta$ isolate karo. Step 3 — ${\sin }^{-1}$ se sine undo karo. ${\sin }^{-1}$ ka jawab hai "kaunse angle ka sine yeh hai?" ${\theta }_1={\sin }^{-1}(0.275)=15.96^{\circ }$ Answer: ${\theta }_1\approx 16.0^{\circ }$.

Step 1: $m=2$: $\sin {\theta }_2=\frac{2\times 550\times 10^{-9}}{2.0\times 10^{-6}}=0.550$ Step 2: ${\theta }_2={\sin }^{-1}(0.550)=33.37^{\circ }$. Answer: ${\theta }_2\approx 33.4^{\circ }$. WHY exactly $2{\theta }_1$ nahi: $\sin \theta$ badhta hai, lekin $\theta$ aur $\sin \theta$ ka relationship curved hai — $90^{\circ }$ ke paas sine flat ho jaata hai, isliye $\sin \theta$ mein barabar jumps ke liye $\theta$ mein zyada bade jumps chahiye jaise hum upar jaate hain. $2\times 16.0^{\circ }=32.0^{\circ }<33.4^{\circ }$. Orders uneven tarike se phailte hain, bade angles par zyada door.

Step 1 — $d$ nikalo. $d=1/(600{\text{mm}}^{-1})=1.667\times 10^{-3}\text{mm}=1.667\times 10^{-6}\text{m}$. Step 2 — order cap karo. $\sin \theta$ kabhi 1 se zyada nahi ho sakta, isliye $m_{\max }=\lfloor \frac{d}{\lambda }\rfloor =\lfloor \frac{1.667\times 10^{-6}}{500\times 10^{-9}}\rfloor =\lfloor 3.33\rfloor =3$ Step 3 — dono sides aur centre ginao. Orders run karte hain $m=-3,-2,-1,0,+1,+2,+3$. $\text{count}=2\times 3+1=7$ Answer: 7 maxima.

Step 1: $\sin {\theta }_3=\frac{3\times 500\times 10^{-9}}{1.667\times 10^{-6}}=0.900$. Step 2: ${\theta }_3={\sin }^{-1}(0.900)=64.16^{\circ }$. Interpretation: $\sin {\theta }_3=0.90<1$, toh yeh order exist karta hai lekin steep $64^{\circ }$ par graze karta hai. Ek hypothetical $m=4$ ke liye $\sin \theta =1.20$ chahiye hoga, jo impossible hai — exactly iseelie floor ne humein 3 par rok diya. Answer: ${\theta }_3\approx 64.2^{\circ }$; $m=4$ forbidden hai ($\sin \theta >1$).

Step 1 — $d$ ke liye rearrange karo. $d\sin \theta =m\lambda$ se: $d=\frac{m\lambda }{\sin \theta }=\frac{3\times 650\times 10^{-9}}{\sin 48.0^{\circ }}$ Step 2 — evaluate karo. $\sin 48.0^{\circ }=0.7431$. $d=\frac{1.95\times 10^{-6}}{0.7431}=2.624\times 10^{-6}\text{m}$ Step 3 — lines per mm. $n=1/d$, mm mein: $d=2.624\times 10^{-3}\text{mm}\Rightarrow n=\frac{1}{2.624\times 10^{-3}}=381\text{lines/mm}$ Answer: $d\approx 2.62\times 10^{-6}\text{m}$, roughly $381\text{lines/mm}$.

Step 1 — violet edge. $\sin {\theta }_V=\frac{1\times 400\times 10^{-9}}{1.6\times 10^{-6}}=0.250$, toh ${\theta }_V={\sin }^{-1}(0.250)=14.48^{\circ }$. Step 2 — red edge. $\sin {\theta }_R=\frac{1\times 700\times 10^{-9}}{1.6\times 10^{-6}}=0.4375$, toh ${\theta }_R={\sin }^{-1}(0.4375)=25.94^{\circ }$. Step 3 — spread. $\Delta \theta ={\theta }_R-{\theta }_V=25.94^{\circ }-14.48^{\circ }=11.46^{\circ }$ WHY red zyada bend karta hai: bada $\lambda$ bada $\sin \theta$ maangta hai taaki $d\sin \theta =\lambda$ satisfy ho, isliye red wider angle par hota hai. Yahi colour-splitting grating spectrometer ka poora point hai. Answer: First-order spectrum $\approx 11.5^{\circ }$ span karta hai (violet $14.5^{\circ }$ par, red $25.9^{\circ }$ par).

Step 1 — har line ka angle, second order ($m=2$). $\sin {\theta }_1=\frac{2\times 589.0\times 10^{-9}}{2.5\times 10^{-6}}=0.47120\Rightarrow {\theta }_1=28.106^{\circ }$ $\sin {\theta }_2=\frac{2\times 589.6\times 10^{-9}}{2.5\times 10^{-6}}=0.47168\Rightarrow {\theta }_2=28.138^{\circ }$ Step 2 — subtract karo. $\Delta \theta ={\theta }_2-{\theta }_1=28.138^{\circ }-28.106^{\circ }=0.032^{\circ }$ Interpretation: $0.6\text{nm}$ ka tiny wavelength gap sirf $\approx 0.03^{\circ }$ kholta hai — isliye doublet resolve karne ke liye ek fine, wide grating aur accha telescope chahiye. Answer: $\Delta \theta \approx 0.032^{\circ }$ (roughly $1.9$ arc-minutes).

Step 1 — order 4 exist karne ki condition. Chahiye $m_{\max }\ge 4$, yaani $d/\lambda \ge 4$, toh $d\ge 4\lambda =4\times 700\times 10^{-9}=2.8\times 10^{-6}\text{m}$ Step 2 — largest $n$ matlab smallest allowed $d$, toh lo $d=2.8\times 10^{-6}\text{m}$: $n=\frac{1}{d}=\frac{1}{2.8\times 10^{-3}\text{mm}}=357\text{lines/mm}$ Toh use karo $n\le 357$ lines/mm (boundary se safe rehne ke liye round down karo). Step 3 — violet ko same grating par check karo. $\frac{d}{{\lambda }_V}=\frac{2.8\times 10^{-6}}{400\times 10^{-9}}=7.0$, toh $m_{\max }(\text{violet})=7\ge 4$. Haan — violet aaram se 4 (aur 7 tak) orders dikhata hai. WHY violet automatically safe hai: violet ka chhota $\lambda$ $d/\lambda$ ko bada banata hai, isliye woh hamesha kam se kam utne hi orders support karta hai jitne red. Red binding constraint hai. Answer: $n\le 357\text{lines/mm}$ ($d=2.8\times 10^{-6}\text{m}$ use karte hue); violet tab 7 tak orders dikhata hai — condition met.

Sabse lambi wavelength ::: kyunki $m_{\max }=\lfloor d/\lambda \rfloor$ $\lambda$ badhne par ghatta hai, isliye red ke orders pehle khatam hote hain.

$d/\lambda$ floor hokar $k$ de? $2k+1$ ::: orders $-k$ se $+k$ tak inclusive, central white maximum ke baare mein symmetric.