Exercises — Diffraction grating — condition for maxima

WHAT: Convert "lines per length" into a spacing. WHY: The grating equation needs $d$ (a distance), never the raw line count. $d=\frac{1}{n}=\frac{1}{300{\text{mm}}^{-1}}=\frac{1}{300}\text{mm}=3.33\times 10^{-3}\text{mm}$ Convert to metres ($1\text{mm}=10^{-3}\text{m}$): $d=3.33\times 10^{-3}\times 10^{-3}\text{m}=3.33\times 10^{-6}\text{m}$ Answer: $d\approx 3.33\times 10^{-6}\text{m}$ (i.e. $3333\text{nm}$).

WHAT: State the extra distance one ray travels. WHY: This is the seed of the whole grating equation. The two rays are parallel, so their only difference is the little segment cut off by dropping a perpendicular from one slit onto the other's ray. That segment is the opposite side of a right triangle whose hypotenuse is $d$ and whose angle is $\theta$: $\Delta =d\sin \theta$ Answer: $\Delta =d\sin \theta$, from the right triangle with hypotenuse $d$ (look at the red segment in the figure above).

Step 1 — pick the order. First order means $m=1$. Step 2 — rearrange for the unknown angle. $\sin {\theta }_1=\frac{m\lambda }{d}=\frac{1\times 550\times 10^{-9}}{2.0\times 10^{-6}}=0.275$ Why rearrange? $\theta$ is what we want, so we isolate $\sin \theta$ first. Step 3 — undo the sine with ${\sin }^{-1}$. ${\sin }^{-1}$ answers "which angle has this sine?" ${\theta }_1={\sin }^{-1}(0.275)=15.96^{\circ }$ 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 not exactly $2{\theta }_1$: $\sin \theta$ grows, but $\theta$ vs $\sin \theta$ is a curved relationship — sine flattens near $90^{\circ }$, so equal jumps in $\sin \theta$ need bigger jumps in $\theta$ as we go up. $2\times 16.0^{\circ }=32.0^{\circ }<33.4^{\circ }$. Orders fan out unevenly, wider apart at large angles.

Step 1 — find $d$. $d=1/(600{\text{mm}}^{-1})=1.667\times 10^{-3}\text{mm}=1.667\times 10^{-6}\text{m}$. Step 2 — cap the order. $\sin \theta$ can never exceed 1, so $m_{\max }=\lfloor \frac{d}{\lambda }\rfloor =\lfloor \frac{1.667\times 10^{-6}}{500\times 10^{-9}}\rfloor =\lfloor 3.33\rfloor =3$ Step 3 — count both sides plus the centre. Orders run $m=-3,-2,-1,0,+1,+2,+3$. $\text{count}=2\times 3+1=7$ 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: $\sin {\theta }_3=0.90<1$, so this order exists but grazes out at a steep $64^{\circ }$. A hypothetical $m=4$ would demand $\sin \theta =1.20$, impossible — that is exactly why the floor cut us off at 3. Answer: ${\theta }_3\approx 64.2^{\circ }$; $m=4$ is forbidden ($\sin \theta >1$).

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

Step 1 — violet edge. $\sin {\theta }_V=\frac{1\times 400\times 10^{-9}}{1.6\times 10^{-6}}=0.250$, so ${\theta }_V={\sin }^{-1}(0.250)=14.48^{\circ }$. Step 2 — red edge. $\sin {\theta }_R=\frac{1\times 700\times 10^{-9}}{1.6\times 10^{-6}}=0.4375$, so ${\theta }_R={\sin }^{-1}(0.4375)=25.94^{\circ }$. Step 3 — the spread. $\Delta \theta ={\theta }_R-{\theta }_V=25.94^{\circ }-14.48^{\circ }=11.46^{\circ }$ WHY red is bent more: larger $\lambda$ needs a larger $\sin \theta$ to keep $d\sin \theta =\lambda$, so red sits at a wider angle. That colour-splitting is the whole point of a grating spectrometer. Answer: first-order spectrum spans $\approx 11.5^{\circ }$ (violet at $14.5^{\circ }$, red at $25.9^{\circ }$).

Step 1 — angle of each line, 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. $\Delta \theta ={\theta }_2-{\theta }_1=28.138^{\circ }-28.106^{\circ }=0.032^{\circ }$ Interpretation: a tiny $0.6\text{nm}$ wavelength gap opens only $\approx 0.03^{\circ }$ — that is why you need a fine, wide grating and a good telescope to resolve the doublet. Answer: $\Delta \theta \approx 0.032^{\circ }$ (about $1.9$ arc-minutes).

Step 1 — condition for order 4 to exist. We need $m_{\max }\ge 4$, i.e. $d/\lambda \ge 4$, so $d\ge 4\lambda =4\times 700\times 10^{-9}=2.8\times 10^{-6}\text{m}$ Step 2 — largest $n$ means smallest allowed $d$, so take $d=2.8\times 10^{-6}\text{m}$: $n=\frac{1}{d}=\frac{1}{2.8\times 10^{-3}\text{mm}}=357\text{lines/mm}$ So use $n\le 357$ lines/mm (round down to be safe of the boundary). Step 3 — check violet on the same grating. $\frac{d}{{\lambda }_V}=\frac{2.8\times 10^{-6}}{400\times 10^{-9}}=7.0$, so $m_{\max }(\text{violet})=7\ge 4$. Yes — violet comfortably shows 4 (and up to 7) orders. WHY violet is safe automatically: violet's smaller $\lambda$ makes $d/\lambda$ larger, so it always supports at least as many orders as red. Red is the binding constraint. Answer: $n\le 357\text{lines/mm}$ (using $d=2.8\times 10^{-6}\text{m}$); violet then shows up to 7 orders — condition met.

The longest wavelength ::: because $m_{\max }=\lfloor d/\lambda \rfloor$ shrinks as $\lambda$ grows, so red runs out of orders first.

$d/\lambda$ floors to $k$? $2k+1$ ::: orders $-k$ through $+k$ inclusive, symmetric about the central white maximum.