The width of the central peak in a single-slit diffraction pattern is \(5.0 \mathrm{mm}\). The wavelength of the light is \(600 \mathrm{nm}\), and the screen is \(2.0 \mathrm{m}\) from the slit. (a) What is the width of the slit? (b) Determine the ratio of the intensity at \(4.5 \mathrm{mm}\) from the center of the pattern to the intensity at the center.

\[\Delta \theta = 2.5 \times 10^{-3}\,rad\] #tag_title#Step 3: Calculate the width of the slit#tag_content#Now we can use the formula for the angular width of the central maximum and the calculated value of \(\Delta \theta\) to find the width of the slit: \[w = \frac{2 \lambda}{\Delta \theta}\] Plugging in the values, we get: \[w = \frac{2 \times 600 \times 10^{-9}\,m}{2.5 \times 10^{-3}\,rad}\] \[w = 4.8 \times 10^{-4}\,m = 0.48\,mm\] So, the width of the slit is \(0.48\,mm\). #b) Finding the ratio of intensities# #tag_title#Step 1: Recall the intensity formula#tag_content#The intensity for the single-slit diffraction pattern at an angle \(\theta\) is given by: \[I(\theta) = I_0 \left[\frac{\sin\left(\frac{\pi w \sin \theta}{\lambda}\right)}{\frac{\pi w \sin \theta}{\lambda}}\right]^2\] where: - \(I(\theta)\) is the intensity at angle \(\theta\) - \(I_0\) is the intensity at the center of the pattern - \(w\) is the width of the slit - \(\lambda\) is the wavelength of light #tag_title#Step 2: Calculate the angle at 4.5mm#tag_content#Using the small angle approximation, we can find the angle at 4.5mm from the center of the pattern: \[\theta = \frac{4.5 \times 10^{-3}\,m}{2.0\,m}\] \[\theta = 2.25 \times 10^{-3}\,rad\] #tag_title#Step 3: Calculate the intensity ratio#tag_content#Now we can use the intensity formula to find the intensity at 4.5mm and the intensity at the center (\(I_0\)). We need to find the ratio of these intensities: \[\frac{I(\theta)}{I_0} = \left[\frac{\sin\left(\frac{\pi w \sin \theta}{\lambda}\right)}{\frac{\pi w \sin \theta}{\lambda}}\right]^2\] Plugging in the values for \(\theta\), \(w\), and \(\lambda\), we get: \[\frac{I(\theta)}{I_0} = \left[\frac{\sin\left(\frac{\pi \times 0.48 \times 10^{-3}\,m \times 2.25 \times 10^{-3}\,rad}{600 \times 10^{-9}\,m}\right)}{\frac{\pi \times 0.48 \times 10^{-3}\,m \times 2.25 \times 10^{-3}\,rad}{600 \times 10^{-9}\,m}}\right]^2\] \[\frac{I(\theta)}{I_0} \approx 0.034\] Therefore, the ratio of the intensity at 4.5mm from the center of the pattern to the intensity at the center is \(\approx 0.034\).