A double slit is illuminated with monochromatic light of wavelength $600.0 \mathrm{nm} .\( The \)m=0\( and \)m=1\( bright fringes are separated by \)3.0 \mathrm{mm}\( on a screen \)40.0 \mathrm{cm}$ away from the slits. What is the separation between the slits? [Hint: Is the small angle approximation justified?]

We are given the following information: - Wavelength of light: \(\lambda = 600 \, \text{nm}\) - Separation between 0th and 1st bright fringes: \(y_{m=1} - y_{m=0} = 3.0 \, \text{mm}\) - Distance between slits and screen: \(L = 40.0 \, \text{cm}\)

Since the given distances are in different units, we need to convert all of them to meters: - \(\lambda = 600 \, \text{nm} = 600 \times 10^{-9} \, \text{m}\) - \(y_{m=1} - y_{m=0} = 3.0 \, \text{mm} = 3.0 \times 10^{-3} \, \text{m}\) - \(L = 40.0 \, \text{cm} = 40 \times 10^{-2} \, \text{m}\)

For the double-slit interference pattern, the formula for the angular separation of bright fringes is: $$\sin{\theta_m} = m\frac{\lambda}{d}$$ where \(\theta_m\) is the angle between the central bright fringe (\(m=0\)) and the \(m\)th bright fringe, \(\lambda\) is the wavelength of the light, and \(d\) is the separation between the slits. For \(m=1\), we have: $$\sin{\theta_1} = \frac{\lambda}{d}$$

The small-angle approximation states that for angles \(\theta\) close to \(0\), \(\sin{\theta} \approx \tan{\theta}\). The angle will be small if the distance between the slits and the screen is much larger than the distance between the fringes. In this case, \(L = 40 \times 10^{-2} \, \text{m}\) and the separation between the fringes is \(3.0 \times 10^{-3} \, \text{m}\), which means that the small-angle approximation is justified. Using the small-angle approximation, we can relate the angle \(\theta_1\) to the distances \(L\) and \(y_{m=1} - y_{m=0}\): $$\tan{\theta_1} \approx \frac{y_{m=1} - y_{m=0}}{L}$$ Since \(\sin{\theta_1} \approx \tan{\theta_1}\), we can write: $$\frac{\lambda}{d} \approx \frac{y_{m=1} - y_{m=0}}{L}$$

Rearranging the equation above to find the separation \(d\), we get: $$d = \frac{\lambda L}{y_{m=1} - y_{m=0}}$$ Now, we can plug in the values we found in Step 2: $$d = \frac{(600 \times 10^{-9} \, \text{m}) \times (40 \times 10^{-2} \, \text{m})}{3.0 \times 10^{-3} \, \text{m}}$$ Calculating the result, we obtain: $$d \approx 8 \times 10^{-6} \, \text{m}$$ Thus, the separation between the slits is approximately \(8 \, \mu\text{m}\).