Q16E - University Physics with Modern Physics

Question

Coherent light of frequency $\mathbf{6}\mathbf{.}\mathbf{32}\mathbf{*}{\mathbf{10}}^{\mathbf{14}}\mathbf{ }\mathit{H}\mathit{z}$ passes through two thin slits and falls on a screen 85.0cm away. You observe that the third bright fringe occurs at $\mathbf{\pm }\mathbf{3}\mathbf{.}\mathbf{11}\mathit{c}\mathit{m}$ on either side of the central bright fringe. (a) How far apart are the two slits? (b) At what distance from the central bright fringe will the third dark fringe occur?

Step-by-Step Solution

Answer

1. The two slits are apart. $\mathbf{3}\mathbf{.}\mathbf{89}\mathbf{*}{\mathbf{10}}^{\mathbf{-}\mathbf{5}}\mathbf{ }\mathit{m}$
2. The third dark fringe will occur at $\mathbf{2}\mathbf{.}\mathbf{60}\mathit{c}\mathit{m}$ from the central bright fringe.

Step 1: Formulas used to solve the question

Position of the $\mathit{m}\mathit{t}\mathit{h}$ bright fringe when the angle is small

                                                      ${\mathit{y}}_{\mathbf{m}}\mathbf{=}\frac{\mathbf{m}\mathbf{\lambda }\mathbf{R}}{\mathbf{d}}$                                      (1)

Location of the dark fringes:

                                                    ${\mathit{y}}_{\mathit{m}}\mathbf{=}\frac{\mathit{m}\mathbf{\left(}{\displaystyle \frac{\mathbf{1}}{\mathbf{2}}}\mathbf{\right)}\mathit{\lambda }\mathit{R}}{\mathit{d}}$                                (2)

Speed of light:

                                                        $\mathit{c}\mathbf{=}\mathit{f}\mathit{\lambda }$                                            (3)

Step 2: Calculate the distance between the two slits

Given: $\mathit{f}\mathbf{=}\mathbf{6}\mathbf{.}\mathbf{32}\mathbf{*}{\mathbf{10}}^{\mathbf{14}}\mathbf{ }\mathit{H}\mathit{z}$   

            $$\begin{gathered} \mathit{R}\mathbf{=}\mathbf{85}\mathit{c}\mathit{m}\mathbf{=}\mathbf{0}\mathbf{.}\mathbf{85}\mathit{m} \\ {\mathit{y}}_{\left(m=3\right)}\mathbf{=}\mathbf{\pm }\mathbf{3}\mathbf{.}\mathbf{11}\mathit{c}\mathit{m}\mathbf{=}\mathbf{\pm }\mathbf{3}\mathbf{.}\mathbf{11}\mathbf{*}{\mathbf{10}}^{\mathbf{-}\mathbf{2}}\mathbf{ }\mathit{m} \\ {\mathit{v}}_{\mathbf{l}\mathbf{i}\mathbf{g}\mathbf{h}\mathbf{t}}\mathbf{=}\mathit{c}\mathbf{=}\mathbf{3}\mathbf{*}{\mathbf{10}}^{\mathbf{8}}\mathbf{ }\mathit{m}\mathbf{/}\mathit{s} \end{gathered}$$

From equation (3),

                                                $\mathit{\lambda }\mathbf{=}\frac{\mathbf{c}}{\mathbf{f}}$                                                       (4)

Now, the Position of the $\mathit{m}\mathit{t}\mathit{h}$ bright fringe when the angle is small

                                                ${\mathit{y}}_{\mathbf{m}}\mathbf{=}\frac{\mathbf{m}\mathbf{\lambda }\mathbf{R}}{\mathbf{d}}$

Solve for $\mathit{d}$ at $\mathit{m}\mathbf{=}\mathbf{3}$

                                                $\mathit{d}\mathbf{=}\frac{\mathbf{3}\mathbf{\lambda }\mathbf{R}}{{\mathbf{y}}_{\mathbf{3}}}$

From equation (4),

                                               $\mathit{d}\mathbf{=}\frac{\mathbf{3}\mathbf{R}}{{\mathbf{y}}_{\mathbf{3}}}\mathbf{*}\frac{\mathbf{c}}{\mathbf{f}}$ 

Plug the given:

                                   $\mathit{d}\mathbf{=}\frac{\mathbf{3}\mathbf{*}\mathbf{0}\mathbf{.}\mathbf{85}}{\mathbf{3}\mathbf{.}\mathbf{11}\mathbf{*}{\mathbf{10}}^{\mathbf{-}\mathbf{2}}}\mathbf{*}\frac{\mathbf{3}\mathbf{.}\mathbf{2}\mathbf{*}{\mathbf{10}}^{\mathbf{8}}}{\mathbf{6}\mathbf{.}\mathbf{32}\mathbf{*}{\mathbf{10}}^{\mathbf{14}}}\mathbf{=}\mathbf{3}\mathbf{.}\mathbf{89}\mathbf{*}{\mathbf{10}}^{\mathbf{-}\mathbf{5}}\mathit{m}$

Step 3: Calculate the distance of the third dark fringe occurfrom the central bright fringe

Here, the Location of the dark fringes is given by

                                                 ${\mathit{y}}_{\mathbf{m}}\mathbf{=}\frac{\mathbf{(}\mathbf{m}\mathbf{+}{\displaystyle \frac{\mathbf{1}}{\mathbf{2}}}\mathbf{)}\mathbf{\lambda }\mathbf{R}}{\mathbf{d}}$

Noting that for third dark fringe is for $\mathit{m}\mathbf{=}\mathbf{2}$.

Hence, $\mathit{m}\mathbf{=}\mathbf{1}$ is the second dark fringe.

So,

                                                          ${\mathit{y}}_{\mathbf{2}}\mathbf{=}\frac{\mathbf{(}\mathbf{2}\mathbf{+}{\displaystyle \frac{\mathbf{1}}{\mathbf{2}}}\mathbf{)}\mathbf{\lambda }\mathbf{R}}{\mathbf{d}}$

From equation (4),

                                                           ${\mathit{y}}_{\mathbf{2}}\mathbf{=}\frac{\mathbf{2}\mathbf{.}\mathbf{5}\mathbf{R}}{\mathbf{d}}\mathbf{*}\frac{\mathbf{c}}{\mathbf{f}}$

Plug the given,

                              ${\mathit{y}}_{\mathbf{2}}\mathbf{=}\frac{\mathbf{2}\mathbf{.}\mathbf{5}\mathbf{*}\mathbf{0}\mathbf{.}\mathbf{85}}{\mathbf{3}\mathbf{.}\mathbf{89}\mathbf{*}{\mathbf{10}}^{\mathbf{-}\mathbf{5}}}\mathbf{*}\frac{\mathbf{3}\mathbf{.}\mathbf{0}\mathbf{*}{\mathbf{10}}^{\mathbf{8}}}{\mathbf{6}\mathbf{.}\mathbf{32}\mathbf{*}{\mathbf{10}}^{\mathbf{14}}}\mathbf{=}\mathbf{0}\mathbf{.}\mathbf{026}\mathit{m}\mathbf{=}\mathbf{2}\mathbf{.}\mathbf{60}\mathit{c}\mathit{m}$

Thus, the two slits are $\mathbf{3}\mathbf{.}\mathbf{89}\mathbf{*}{\mathbf{10}}^{\mathbf{-}\mathbf{5}}\mathbf{ }\mathit{m}$ apart.The third dark fringe will occur at $\mathbf{2}\mathbf{.}\mathbf{60}\mathit{c}\mathit{m}$ from the central bright fringe.