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?

Short 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-by-step solution

Formula for the distance of the bright fringe

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

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 $\mathbf{2}\mathbf{.}\mathbf{53}\mathit{m}\mathit{m}$.