Q52P - University Physics with Modern Physics

Question

In your summer job at an optics company, you are asked to measure the wavelength l of the light that is produced by a laser. To do so, you pass the laser light through two narrow slits that are separated by a distance d. You observe the interference pattern on a screen that is 0.900 m from the slits and measure the separation ∆y between adjacent bright fringes in the portion of the pattern that is near the center of the screen. Using a microscope, you measure d. But both ∆y and d are small and difficult to measure accurately, so you repeat the measurements for several pairs of slits, each with a different value of d. Your results are shown in Fig. P35.52, where you have plotted ∆y versus 1>d. The line in the graph is the best-fit straight line for the data.
(a) Explain why the data points plotted this way fall close to a straight line. (b) Use Fig. P35.52 to calculate l.

Step-by-Step Solution

Answer

1. Since $\mathrm{y}\infty \frac{1}{\mathrm{d}}$ we get close to a straight line
2. The wavelength is 617nm

Step 1: Important Concepts

Position of bright fringe is given by 

                                                    ${\mathit{y}}_{\mathbf{m}}\mathbf{=}\frac{\mathbf{m}\mathbf{\lambda }\mathbf{D}}{\mathbf{2}}$

Where m is an integer, $\mathit{\lambda }$ is the wavelength of the light, D is the distance of the screen and d is the distance between slits.

Step 2: Application

We know that the position of the $m^{th}$ fringe is given by 

                                                   $y_m=\frac{m\lambda D}{2d}$

And the position of the ${\left(\mathrm{m}+1\right)}^{\mathrm{th}}$ bright fringe is given by 

                                                       ${\mathrm{y}}_{\mathrm{m}}=\frac{\left(\mathrm{m}+1\right)\mathrm{\lambda D}}{2\mathrm{d}}$

Distance between the two fringes is given by

                                          $$\begin{gathered} ∆\mathrm{y}=\frac{(\mathrm{m}+1)\mathrm{\lambda D}}{2\mathrm{d}}-\frac{\mathrm{m\lambda D}}{2\mathrm{d}} \\ ∆\mathrm{y}=\frac{\mathrm{\lambda D}}{2\mathrm{d}} \\ ∆\mathrm{y}=\frac{\mathrm{\lambda R}}{\mathrm{d}} \end{gathered}$$

It is easy to conclude from this that $∆y$ is proportional to 1/d since both $\lambda$ and R are constant 

                                         $∆\mathrm{y}\infty \frac{1}{\mathrm{d}}$

This explains why the data points are plotted as straight line.

Step 3: Finding wavelength

Assume 

$\frac{1}{\mathrm{d}}=\mathrm{x}$

Plug this into the equation 

                                          $$\begin{gathered} ∆\mathrm{y}=\mathrm{\lambda R}\frac{1}{\mathrm{d}} \\ ∆\mathrm{y}=\mathrm{\lambda Rx} \end{gathered}$$                               

Divide both sides by x

We get

                                            $\frac{∆\mathrm{y}}{∆\mathrm{x}}=\mathrm{\lambda R}$                                                 

 

We know that the slope of a straight line is given by 

                                          $$\begin{gathered} \mathrm{slope}=\frac{∆\left(∆\mathrm{y}\right)}{∆\left({\displaystyle \frac{1}{\mathrm{d}}}\right)} \\ \mathrm{slope}=\frac{∆\mathrm{y}}{∆\mathrm{x}} \end{gathered}$$         

 

Plugging in values from the graph

                                                 $$\begin{gathered} \frac{\mathit{∆}\mathit{y}}{\mathit{∆}\mathit{x}}\mathit{=}\frac{\left(5.50-0.50\right)\mathit{m}\mathit{m}}{\left(10.0-1.0\right)\mathit{m}{\mathit{m}}^{\mathit{-}\mathit{1}}} \\ \frac{\mathit{∆}\mathit{y}}{\mathit{∆}\mathit{x}}\mathit{=}\frac{\mathit{5}}{\mathit{9}}m{m}^{\mathit{2}} \end{gathered}$$

Plug this into 

                                        $\frac{5}{9}={\left({10}^{-3}\right)}^2=\mathrm{\lambda R}$                     

Solve for $\lambda$

                                       $$\begin{gathered} \lambda =\frac{{\displaystyle \frac{5}{9}}{\left({10}^{-3}\right)}^2}{0.90} \\ \lambda =617nm \end{gathered}$$

                                           

Hence, The wavelength is 617nm.