Chapter 25, Problem 36P

To determine

Find the separation between the slits.

Answer to Problem 36P

The separation between the slits is $\underset{\_}{0.15\text{mm}}$.

Explanation of Solution

The schematic diagram of the slit and the interference process is shown in figure 1 below.

The angle of interference of both red and violet lights is very small. So $\sin \theta \approx \theta$. Also the distance $s$ separating the violet first order maximum is very small compared to the distance to the screen $D$. Thus $s\approx L\theta$.

Write the equation to find the wavelength of red light.

    $\begin{array}{c}{\lambda }_{red}=d\sin {\theta }_{red}\\ =d{\theta }_{red}\end{array}$                                                                                                      (I)

Here, ${\lambda }_{red}$ is the wavelength of red light, $d$ is the slit separation, ${\theta }_{red}$ is the angle of interference

Write the equation to find the wavelength of violet light.

    $\begin{array}{c}{\lambda }_{violet}=d\sin {\theta }_{violet}\\ =d{\theta }_{violet}\end{array}$                                                                                                      (II)

Here, ${\lambda }_{violet}$ is the wavelength of violet light, $d$ is the slit separation, ${\theta }_{violet}$ is the angle of interference

Write the equation to find the angle of interference.

    $\theta =\frac{s}{L}$                                                                                                                     (III)

Divide equations (I) and (II) and find ${\theta }_{violet}$

    $\begin{array}{c}\frac{d{\theta }_{red}}{d{\theta }_{violet}}=\frac{{\lambda }_{red}}{{\lambda }_{violet}}\\ {\lambda }_{violet}=\frac{{\lambda }_{violet}}{{\lambda }_{red}}{\theta }_{red}\end{array}$                                                                                                (IV)

Write the equation to find the difference of ${\theta }_{red}$ and ${\theta }_{violet}$

    $\begin{array}{c}{\theta }_{red}-{\theta }_{violet}={\theta }_{red}-\frac{{\lambda }_{violet}}{{\lambda }_{red}}{\theta }_{red}\\ \approx \frac{s}{L}\end{array}$                                                                                    (V)

Rewrite equation (V) to get ${\theta }_{red}$

    ${\theta }_{red}=\frac{s}{L}{\left(1-\frac{{\lambda }_{violet}}{{\lambda }_{red}}\right)}^{-1}$                                                                                            (VI)

Multiply equation (VI) by $d$ and simplify equation to get $d$

    $\begin{array}{c}d{\theta }_{red}=d\frac{s}{L}\left(1-\frac{{\lambda }_{violet}}{{\lambda }_{red}}\right)={\lambda }_{red}\\ d=\frac{{\lambda }_{red}L}{s}\left(1-\frac{{\lambda }_{violet}}{{\lambda }_{red}}\right)\\ =\frac{L\left({\lambda }_{red}-{\lambda }_{violet}\right)}{s}\end{array}$                                                                                     (VII)

Conclusion

Substitute $3.40\text{m}$ for $L$ , $700\text{nm}$ for ${\lambda }_{red}$ , $400\text{nm}$ for ${\lambda }_{violet}$ , $7.00\text{mm}$ for $s$ in equation (VII) to get $d$

    $\begin{array}{c}d=\frac{\left(3.40\text{m}\right)\left(700\text{nm}\left(\frac{1\text{m}}{{10}^9\text{nm}}\right)-400\text{nm}\left(\frac{1\text{m}}{{10}^9\text{nm}}\right)\right)}{7.00\text{mm}\left(\frac{1\text{m}}{{10}^3\text{mm}}\right)}\\ =0.15\text{mm}\end{array}$

Therefore, the separation between the slits is $\underset{\_}{0.15\text{mm}}$.