91影视

The luminous fringe arises when the crest of one wave coincides with the crest of another. The dark fringe occurs when the trough of one wave coincides with the trough of another, tends to result in dark fringes.

The distance between two $m^{th}$order brilliant fringes will be given by in the preceding experiment.

$x\left(m\right)=2Y_m=2L\tan 鈦�\left({\sin }^{鈭�1}鈦�\left(\frac{m位}{d}\right)\right)$

We would identify which integer $m$yields the greatest localid="1649156590466" $x$, lower than $2$, because our screen has a specified width of localid="1649156593812" $2m$. Set the equation to localid="1649156597913" $2$and then choose the lowest integer that is less than the answer. To put it another way,

$2=2L\tan 鈦�\left({\sin }^{鈭�1}鈦�\left(\frac{m位}{d}\right)\right)$

We obtain by substitutinglocalid="1649156604840" $L=2$and modifying

localid="1649096629498" $$\begin{gathered} 0.5=\tan 鈦�\left({\sin }^{鈭�1}鈦�\left(\frac{m位}{d}\right)\right) \\ 鈬�{\sin }^{鈭�1}鈦�\left(\frac{m位}{d}\right)={26.6}^{鈭�}. \end{gathered}$$

Considering this, we may calculate $m$as the limiting.

$$\begin{gathered} \frac{m位}{d}=\sin 鈦�\left({26.6}^{鈭�}\right) \\ 鈬�m=\frac{0.4472脳d}{位} \end{gathered}$$

Knowing that the grating constant is $500$lines per millimetre, we can get $d=2渭m$. We now have the wavelength and can calculate the maximum$m$:

$m=\frac{0.4472脳2脳{10}^{鈭�6}}{5.1脳{10}^{鈭�7}}=\underset{\_}{1.7鈥�鈥�}$

Consequently, this means that on each side of the centre maximum, only one diffracted ray will be shown on the screen. As a result, there are three dazzling fringes in total.