Q14E - University Physics with Modern Physics

Question

Coherent light that contains two wavelengths, 660nm (red) and 470nm (blue), passes through two narrow slits that are separated by 0.300mm. Their interference pattern is observed on a screen 4.00m from the slits. What is the distance on the screen between the first-order bright fringes for the two wavelengths?

Step-by-Step Solution

Answer

The distance on the screen between the first-order bright fringes for the two wavelengths is $\mathbf{2}\mathbf{.}\mathbf{53}\mathit{m}\mathit{m}$.

Step 1: Formula for the distance of the bright fringe

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

Step 2: Determining the distance on the screen between the first-order bright fringes for the two wavelengths

Given: ${\mathit{\lambda }}_{\mathbf{r}\mathbf{e}\mathbf{d}}\mathbf{=}\mathbf{660}\mathit{n}\mathit{m}\mathbf{=}\mathbf{660}\mathbf{*}{\mathbf{10}}^{\mathbf{-}\mathbf{9}}\mathit{m}$

            $$\begin{gathered} {\mathit{\lambda }}_{\mathbf{blue}}\mathbf{=}\mathbf{470}\mathit{n}\mathit{m}\mathbf{=}\mathbf{470}\mathbf{*}{\mathbf{10}}^{\mathbf{-}\mathbf{9}}\mathbf{ }\mathit{m} \\ \mathit{d}\mathbf{=}\mathbf{0}\mathbf{.}\mathbf{300}\mathit{m}\mathit{m}\mathbf{=}\mathbf{0}\mathbf{.}\mathbf{300}\mathbf{*}{\mathbf{10}}^{\mathbf{-}\mathbf{3}}\mathbf{ }\mathit{m} \\ \mathit{R}\mathbf{=}\mathbf{4}\mathbf{.}\mathbf{0}\mathit{m} \\ \mathit{m}\mathbf{=}\mathbf{1} \end{gathered}$$

Assume that the angles of both lights are small, so the distance can be given as:

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

The distance between them can be given as:

                                                              $\mathbf{∆}\mathit{y}\mathbf{=}{\mathit{y}}_{\mathbf{r}\mathbf{e}\mathbf{d}}\mathbf{-}{\mathit{y}}_{\mathbf{b}\mathbf{l}\mathbf{u}\mathbf{e}}$

Plug the first formula into the last one,

                                                          $$\begin{gathered} \mathbf{∆}\mathit{y}\mathbf{=}\frac{\mathbf{m}{\mathbf{\lambda }}_{\mathbf{r}\mathbf{e}\mathbf{d}}\mathbf{R}}{\mathbf{d}}\mathbf{-}\frac{\mathbf{m}{\mathbf{\lambda }}_{\mathbf{b}\mathbf{l}\mathbf{u}\mathbf{e}}\mathbf{R}}{\mathbf{d}} \\ \mathbf{ }\mathbf{∆}\mathit{y}\mathbf{=}\frac{\mathbf{m}\mathbf{R}}{\mathbf{d}}\left({\lambda }_{red}-{\lambda }_{blue}\right) \end{gathered}$$

Plug the given,

                             $\mathbf{∆}\mathit{y}\mathbf{=}\frac{\mathbf{1}\mathbf{.}\mathbf{0}\mathbf{*}\mathbf{4}\mathbf{.}\mathbf{0}}{\mathbf{0}\mathbf{.}\mathbf{300}\mathbf{*}{\mathbf{10}}^{\mathbf{-}\mathbf{3}}}\left(660*{10}^{-9}-470*{10}^{-9}\right)\mathbf{=}\mathbf{2}\mathbf{.}\mathbf{53}\mathbf{*}{\mathbf{10}}^{\mathbf{-}\mathbf{3}}\mathit{m}\mathbf{=}\mathbf{2}\mathbf{.}\mathbf{53}\mathit{m}\mathit{m}$

Thus, the distance on the screen between the first-order bright fringes for the two wavelengths is $\mathbf{2}\mathbf{.}\mathbf{53}\mathit{m}\mathit{m}$.