91影视

The chart below shows that wave 1 travels an incremental $\sin 鈦�\left(蠒\right)=\frac{r_1}{d}$$r_1$distance prior approaching the reflection spot, and since $\sin 鈦�\left(蠒\right)=\frac{r_1}{d}$, we may write $r_1=d\sin 鈦�\left(蠒\right)$. Wave $2$behaves similarly, with an added distance $r_2$from of the current point, where $r_2=d\sin 鈦�\left(胃\right)$. As a result, the difference in path length is

$$\begin{gathered} L=r_2鈭�r_1 \\ =d\sin 鈦�\left(胃\right)鈭�d\sin 鈦�\left(蠒\right) \\ =d(\sin 鈦�(胃)鈭�\sin 鈦�(蠒\left)\right) \end{gathered}$$

The phase shift must be equal to $m位$for diffraction to occur; comparing $m位$with $\mathrm{螖}r$from part (a) yields

$d\sin 鈦�\left({胃}_m\right)鈭�d\sin 鈦�\left(蠒\right)=m位$

Hence

$$\begin{gathered} d\sin 鈦�\left({胃}_m\right)=m位+d\sin 鈦�\left(蠒\right) \\ m=鈥�鈭�2,鈭�1,0,1,2,鈥� \end{gathered}$$

So we want to prove that ${胃}_0=蠒\text{when}m=0$let's replace $m$in the formula with zero to see what we get.

$\begin{array}{c}d\sin 鈦�\left({胃}_0\right)=0+d\sin 鈦�\left(蠒\right)\end{array}$

$\sin 鈦�\left({胃}_0\right)=\sin 鈦�\left(蠒\right)$

${胃}_0=蠒$

Let us just start by determining the spacing between the grating's reflecting lines.

$$\begin{gathered} d=\frac{1脳{10}^{鈭�3}\mathrm{m}}{700} \\ =1.4285脳{10}^{鈭�6}\mathrm{m} \\ =1428.5\mathrm{nm} \end{gathered}$$

To isolate ${胃}_m$, rearranging the equation we found in step (b) yields

${胃}_m={\sin }^{鈭�1}鈦�\left(\frac{m位}{d}+\sin 鈦�蠒\right)$

$胃\text{are}胃=0,胃鈮�{90}^{鈭�}\text{and}胃鈮�鈭�{90}^{鈭�}$are all forbidden values for $胃$So let's start by changing the value of $m$and to see what angle $胃$matches to each $m$

$m=0$

${胃}_0={\sin }^{鈭�1}鈦�\left(0+\sin 鈦�\left({40}^{鈭�}\right)\right)={40}^{鈭�}$

$m=1$

${胃}_1={83.12}^{鈭�}$

$m=-1$

$\begin{array}{c}{胃}_{鈭�1}={\sin }^{鈭�1}鈦�\left(\frac{鈭�500\mathrm{nm}}{1428.5\mathrm{nm}}+\sin 鈦�\left({40}^{鈭�}\right)\right)\\ \end{array}$

${胃}_{鈭�1}={17.02}^{鈭�}$

$m=2$

${胃}_2={\sin }^{鈭�1}鈦�\left(\frac{2脳500\mathrm{nm}}{1428.5\mathrm{nm}}+\sin 鈦�\left({40}^{鈭�}\right)\right)$

The value within the ${\sin }^{鈭�1}鈦�\text{is}1.342$, which is outside the domain of the ${\sin }^{鈭�1}鈦�\left(x\right)$function, which is $鈭�1鈮�x鈮�1$As a consequence, $胃2$is ill defined, and finding$胃$for any $m$greater than $2$is worthless.

$\begin{array}{c}{胃}_{鈭�2}={\sin }^{鈭�1}鈦�\left(\frac{鈭�2脳500\mathrm{nm}}{1428.5\mathrm{nm}}+\sin 鈦�\left({40}^{鈭�}\right)\right)\end{array}$

${胃}_{鈭�2}=鈭�{3.28}^{鈭�}$

${胃}_{鈭�3}={\sin }^{鈭�1}鈦�\left(\frac{鈭�3脳500\mathrm{nm}}{1428.5\mathrm{nm}}+\sin 鈦�\left({40}^{鈭�}\right)\right)$

${胃}_{鈭�3}=鈭�{24.03}^{鈭�}$

${胃}_{鈭�4}={\sin }^{鈭�1}鈦�\left(\frac{鈭�4脳500\mathrm{nm}}{1428.5\mathrm{nm}}+\sin 鈦�\left({40}^{鈭�}\right)\right)$

${胃}_{鈭�4}=鈭�{49.22}^{鈭�}$

${胃}_{鈭�5}={\sin }^{鈭�1}鈦�\left(\frac{鈭�5脳500\mathrm{nm}}{1428.5\mathrm{nm}}+\sin 鈦�\left({40}^{鈭�}\right)\right)$

The value from the inside of the ${\sin }^{鈭�1}$function for $m=-5$is $(-1.1073)$, which is beyond the domain of the ${\sin }^{鈭�1}\left(x\right)$function. As a result, $胃-5$is indefinite, and there is no need for look for $胃$for any$m$less than $-5$

To construct the proper graph, keep in mind that negatives $胃$values belong to ratios on the left side of the horizontal, while positive values correspond to angles on the correct side.