Q12E - University Physics with Modern Physics

Question

Coherent light with wavelength $\mathbf{400}$ nm passes through two very narrow slits that are separated by  $\mathbf{0}\mathbf{.}\mathbf{200}$nm, and the interference pattern is observed on a screen $\mathbf{4}\mathbf{.}\mathbf{00}$nm from the slits. (a) What is the width (in nm) of the central interference maximum? (b) What is the width of the first-order bright fringe?

Step-by-Step Solution

Answer

(a) The width of the central interference maximum is 8.0 mm.

(b) The width of the first-order bright fringe is 8.0 mm.

Step 1: Formula for the distance between any two dark fringes

                            $\mathbf{△}\mathbf{y}\mathbf{=}{\mathbf{y}}_{\mathbf{m}\mathbf{+}\mathbf{1}}\mathbf{-}{\mathbf{y}}_{\mathbf{m}}\mathbf{=}\frac{\left(\left|m+1\right|{\displaystyle \frac{1}{2}}\right)\mathbf{\lambda R}}{\mathbf{d}}\mathbf{-}\frac{\left(m+{\displaystyle \frac{1}{2}}\right)\mathbf{\lambda R}}{\mathbf{d}}$

Step 2: Calculation of the width of the central interference maximum

Given: 

             $$\begin{gathered} \mathrm{\lambda }=400\mathrm{nm} \\ \mathrm{d}=0.200\mathrm{nm}=0.200*{10}^{-3} \mathrm{m} \\ \mathrm{R}=4\mathrm{m} \end{gathered}$$ 

               

    (a) For double-slit destructive interference,

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

 

And hence, the distance between any two dark fringes is given as:

                                $$\begin{gathered} △\mathrm{y}={\mathrm{y}}_{\mathrm{m}+1}-{\mathrm{y}}_{\mathrm{m}}=\frac{(\left|\mathrm{m}+1\right|+{\displaystyle \frac{1}{2}})\mathrm{\lambda R}}{\mathrm{d}}-\frac{(\mathrm{m}+{\displaystyle \frac{1}{2}})\mathrm{\lambda R}}{\mathrm{d}} \\ \Rightarrow △\mathrm{y}=\frac{(\mathrm{m}1.5)\mathrm{\lambda R}}{\mathrm{d}}-\frac{(\mathrm{m}+{\displaystyle \frac{1}{2}})\mathrm{\lambda R}}{\mathrm{d}} \\ \Rightarrow △\mathrm{y}=(\mathrm{m}+1.5-\mathrm{m}-\frac{1}{2})\frac{\mathrm{\lambda R}}{\mathrm{d}} \\ \Rightarrow △y=\frac{\mathrm{\lambda R}}{\mathrm{d}} \end{gathered}$$

                                                           (1)

    Plug the given into (1),

                                          $△\mathrm{y}=\frac{400*{10}^{-9}*4.0}{0.200*{10}^{-3}}=8.0\mathrm{mm}$

  

Step 3: Calculation for the width of the first-order bright fringe

Since the width of the first-order bright fringe is the distance between the two dark fringes surrounding it, so the same answer applies here.

Hence, 

                                                                             $△\mathrm{y}=8.0\mathrm{mm}$

Thus, the width of the central interference maximum is $8.0 \mathrm{mm}$  and the width of the first-order bright fringe is also $8.0\mathrm{mm}$