Two thin parallel slits that are $\mathbf{0}\mathbf{.}\mathbf{0116}$mm apart are illuminated by a laser beam of wavelength $\mathbf{585}$mm. (a) On a very large distant screen, what is the total number of bright fringes (those indicating complete constructive interferences), including the central fringe and those on both sides of it? Solve this problem without calculating all the angles! (Hint: What is the largest that$\mathbf{sin\theta }$can be? What does this tell you is the largest value of$\mathit{m}$?) (b) At what angle, relative to the original direction of the beam, will the fringe that is most distant from the central bright fringe occur?

Distance between slits:

$$\begin{gathered} \mathrm{d}=0.0116\mathrm{mm} \\ =0.0116\times {10}^{-3}\mathrm{m} \end{gathered}$$

Wavelength:

role="math" localid="1663933162734" $\begin{array}{l}\mathrm{\lambda }=585\mathrm{nm}\\ =585\times {10}^{-9}\mathrm{m}\end{array}$

The maximum angle for the farthest bright from the central bright fringe is${\mathbf{90}}^{\mathbf{°}}$.

(a) First, find the maximum angle of this fringe.

The maximum angle for the farthest bright fringe from the central bright fringe is .

Hence,

$\mathrm{d}\sin {90}^{\circ }={\mathrm{m}}_{{\max }^{\mathrm{\lambda }}}$

Solve for m,

$m_{\max }=\frac{d}{\lambda }$

Plug the given,

localid="1663933487540" ${\mathrm{m}}_{\max }=\frac{0.0116\times {10}^{-3}}{585\times {10}^{-9}\mathrm{m}}=19.8$

Now, $m$must be the integer number. This means that the farthest bright fringe range is at

${\mathrm{m}}_{\max }=19$

Hence, there are $19$bright fringes above the central bright fringe, plus $19$fringes below the central bright fringe and the central bright fringe itself.

So, the total number of bright fringes is

$$\begin{gathered} \mathrm{N}=19+19+1 \\ =39 \end{gathered}$$

(b) Solve (1) for $\theta$to find its angle.

$$\begin{gathered} \mathrm{sin\theta }=\frac{{{\mathrm{m}}_{\max }}^{\mathrm{\lambda }}}{\mathrm{d}} \\ \mathrm{\theta }={\sin }^{-1}\left(\frac{{{\mathrm{m}}_{\max }}^{\mathrm{\lambda }}}{\mathrm{d}}\right) \end{gathered}$$

$$\begin{gathered} \theta ={\sin }^{-1}\left(\frac{19\times 585\times {10}^{-9}}{0.0116\times {10}^{-3}}\right) \\ =73.4 \end{gathered}$$

Hence, the total number of bright fringes are$39$and $73.4^{°}$is the angular separation of the farthest fringe from central fringe.