Q15E - University Physics with Modern Physics

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?

Step-by-Step Solution

Answer

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

Step 1: Formulas used to solve the question

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

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

For dark fringes,

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

Step 2: 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 $\mathit{m}\mathbf{=}\mathbf{1}$;

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

Step 3: 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 $\mathbf{1200}\mathit{n}\mathit{m}$.