91影视

To begin, remember that we won't utilise the little angle approximation in optical device issues since the angle in these situations is never small. Now we must determine the gap between the two brilliant fringes, each of which is on the other side of the centre bright fringe. We'll measure the space between the two fringes, and also the spacing between the two fringes are double that measurement "(due to symmetry)". the shape of the standards for colorful fringes is

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

$\left({胃}_1\right)$is what we've been looking for. Thus

${胃}_1={\sin }^{鈭�1}鈦�\left(\frac{位}{d}\right)$

Given that the optical device contains ,the space between any two successive slits is

$$\begin{gathered} d=\frac{1\mathrm{mm}}{600} \\ d=1.66脳{10}^{鈭�6}\mathrm{m}. \end{gathered}$$

We can use the subsequent relationship to compute the space of the primary bright fringe from the centre maximum now that we all know the angle of the primary bright fringe

$\begin{array}{l}{y}_1=L\tan 鈦�\left({胃}_1\right)\\ y_1=\left(2\mathrm{m}\right)脳\tan 鈦�\left({17.9}^{鈭�}\right)\\ y_1=0.645\mathrm{m}.\end{array}$

As a result, the gap between the two brilliant fringes with $m=1$is

$$\begin{gathered} \mathrm{螖}y=2y_1 \\ 鈭�y=1.29\mathrm{m}. \end{gathered}$$

Equation as

$\begin{array}{l}\sin 鈦�\left({胃}_m\right)=\frac{m位}{d}\\ \left|\sin 鈦�\left({胃}_m\right)\right|鈮�1\end{array}$

The maximum no of bright fringes of central increases as

$$\begin{gathered} m_{max}=\frac{d}{位} \\ {m}_{max}=\frac{1.66脳{10}^{鈭�6}\mathrm{m}}{510脳{10}^{鈭�9}\mathrm{m}} \\ {m}_{max}=3.25. \end{gathered}$$

So,the number of fringes on screen is

$$\begin{gathered} n=(2脳3)+1 \\ n=7. \end{gathered}$$