Question

Coherent light that contains two wavelengths, 600nm (red) and 470nm (blue), passes through two narrow slits that are separated by 0.300mm. Their interference pattern is observed on a screen 3.00m from the slits. The first-order bright fringe is at 4.84mm from the center of the central bright fringe. For what wavelength of light will the first-order dark fringe be observed at this same point on the screen?

Short answer

The wavelength of light is$\mathbf{1200}\mathit{n}\mathit{m}$.

Step-by-step solution

Formulas used to solve the question

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

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

For dark fringes,

$\mathit{d}\mathit{s}\mathit{i}\mathit{n}\mathit{\theta }\mathbf{=}\mathbf{(}\mathit{m}\mathbf{+}\frac{\mathbf{1}}{\mathbf{2}}\mathbf{)}\mathit{\lambda }$

Calculate the separation between the two slits

Given: ${\mathit{\lambda }}_{\mathbf{1}}\mathbf{=}\mathbf{600}\mathit{n}\mathit{m}\mathbf{=}\mathbf{600}\mathbf{*}{\mathbf{10}}^{\mathbf{-}\mathbf{9}}\mathbf{}\mathit{m}$

$$\begin{gathered} \mathit{R}\mathbf{=}\mathbf{3}\mathbf{.}\mathbf{00}\mathit{m} \\ {\mathit{y}}_{\left(m=1\right)}\mathbf{=}\mathbf{4}\mathbf{.}\mathbf{84}\mathit{m}\mathit{m}\mathbf{=}\mathbf{4}\mathbf{.}\mathbf{84}\mathbf{*}{\mathbf{10}}^{\mathbf{-}\mathbf{3}}\mathbf{}\mathit{m} \end{gathered}$$

The first bright fringe for$\mathit{m}\mathbf{=}\mathbf{1}$ is at$\mathbf{4}\mathbf{.}\mathbf{84}\mathit{m}\mathit{m}$ from the central bright fringe when the light of$\mathbf{600}\mathit{n}\mathit{m}$ wavelength is used.

So, first find the separation between these two slits

The position of the mth bright fringe when the angle is small is given by

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

Solve for :

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

Plug the given, when $\mathit{m}\mathbf{=}\mathbf{1}$;

$\mathit{d}\mathbf{=}\frac{\mathbf{1}\mathbf{.}\mathbf{0}\mathbf{*}\mathbf{600}\mathbf{*}{\mathbf{10}}^{\mathbf{-}\mathbf{9}}\mathbf{*}\mathbf{3}\mathbf{.}\mathbf{0}}{\mathbf{4}\mathbf{.}\mathbf{84}\mathbf{*}{\mathbf{10}}^{\mathbf{-}\mathbf{3}}}\mathbf{=}\mathbf{3}\mathbf{.}\mathbf{72}\mathbf{*}{\mathbf{10}}^{\mathbf{-}\mathbf{4}}\mathbf{}\mathit{m}$

Calculate the wavelength

Since$\mathit{R}$ is much greater than$\mathit{d}$ , so using the formula of small angles is acceptable.

Now, for dark fringes

$\mathit{d}\mathit{s}\mathit{i}\mathit{n}\mathit{\theta }\mathbf{=}\mathbf{(}\mathit{m}\mathbf{+}\frac{\mathbf{1}}{\mathbf{2}}\mathbf{)}\mathit{\lambda }$

At the first-order dark fringe, $\mathit{m}\mathbf{=}\mathbf{0}$

$$\begin{gathered} \mathbf{\Rightarrow }\mathit{d}\mathit{s}\mathit{i}\mathit{n}\mathit{\theta }\mathbf{=}\mathbf{(}\mathbf{0}\mathbf{+}\frac{\mathbf{1}}{\mathbf{2}}\mathbf{)}{\mathit{\lambda }}_{\mathbf{2}} \\ \mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{\Rightarrow }\mathbf{sin}\mathit{\theta }\mathbf{=}\frac{{\mathbf{\lambda }}_{\mathbf{2}}}{\mathbf{2}\mathbf{d}} \end{gathered}$$

Noting that the angle of the first order bright fringe is the same angle as this case since both fringes are located at the same position.

Therefore,

$$\begin{gathered} \mathit{s}\mathit{i}\mathit{n}\mathit{\theta }\mathbf{=}\frac{{\mathbf{\lambda }}_{\mathbf{2}}}{\mathbf{2}\mathbf{d}}\mathbf{=}\frac{\mathbf{m}{\mathbf{\lambda }}_{\mathbf{1}}}{\mathbf{d}} \\ \mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{\Rightarrow }\frac{{\mathbf{\lambda }}_{\mathbf{2}}}{\mathbf{2}\mathbf{d}}\mathbf{=}\frac{\mathbf{1}{\mathbf{\lambda }}_{\mathbf{1}}}{\mathbf{d}} \\ \mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{\Rightarrow }{\mathit{\lambda }}_{\mathbf{2}}\mathbf{=}\mathbf{2}{\mathit{\lambda }}_{\mathbf{1}} \end{gathered}$$

Plug the given,

$$\begin{gathered} {\mathit{\lambda }}_{\mathbf{2}}\mathbf{=}\mathbf{2}\mathbf{*}\mathbf{600}\mathbf{*}{\mathbf{10}}^{\mathbf{-}\mathbf{9}}\mathbf{=}\mathbf{1200}\mathbf{*}{\mathbf{10}}^{\mathbf{-}\mathbf{9}\mathbf{m}} \\ \mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{=}\mathbf{1200}\mathit{n}\mathit{m} \end{gathered}$$

Thus, the wavelength of light is $\mathbf{1200}\mathit{n}\mathit{m}$.