Fifty-one narrow slits are equally spaced and separated by \(0.10 \mathrm{mm}\). The slits are illuminated by blue light of wavelength \(400 \mathrm{nm} .\) What is angular position of the twenty-fifth secondary maximum? What is its peak intensity in comparison with that of the primary maximum?

We can use the equation for angular positions of maxima in a multiple-slit interference pattern: \[m \lambda = d \sin(\theta)\] where \(m\) is the order of the maximum, \(\lambda\) is the wavelength of light, \(d\) is the separation between slits, and \(\theta\) is the angular position of the maximum. In this case, the twenty-fifth secondary maximum corresponds to \(m=25\), \(\lambda = 400 \text{ nm}\) and \(d = 0.10 \text{ mm}\). First, we need to convert the given values into the same units (meters). The wavelength \(\lambda = 400 \text{ nm} = 400 \times 10^{-9} \text{ m}\) and the separation between slits \(d = 0.10 \text{ mm} = 0.10 \times 10^{-3} \text{ m}\). Now, we plug the values into the formula and solve for the angular position (\(\theta\)): \[25(400 \times 10^{-9} \text{ m}) = (0.10 \times 10^{-3} \text{ m}) \sin(\theta)\] Solve for \(\theta\): \[\theta = \arcsin \left( \frac{25(400 \times 10^{-9} \text{ m})}{0.10 \times 10^{-3} \text{ m}} \right)\] \[\theta \approx 1.733^\circ\] Therefore, the angular position of the twenty-fifth secondary maximum is approximately \(1.733^\circ\).

We will use the formula for the intensity ratio of secondary maxima to the primary maximum in a multiple-slit interference pattern. This formula is given by: \[\frac{I_{m}}{I_{0}} = \frac{\sin^2(\beta)}{N^2 \sin^2(\frac{\beta}{N})}\] where \(I_{m}\) is the intensity of the secondary maximum, \(I_{0}\) is the intensity of the primary maximum, \(\beta\) is a term related to the path differences, given by \(\beta = \frac{\pi d}{\lambda}\sin(\theta)\), and \(N\) is the total number of slits. In our case, \(m = 25\), \(\theta \approx 1.733^\circ\), \(N=51\), and \(\beta = \frac{\pi(0.10 \times 10^{-3} \text{ m})}{400 \times 10^{-9} \text{ m}}\sin(1.733^\circ)\). Calculate \(\beta\): \[\beta \approx \frac{\pi(0.10 \times 10^{-3} \text{ m})}{400 \times 10^{-9} \text{ m}}\sin(1.733^\circ) \approx 15.41\] Now, we can plug the values back into the intensity ratio formula: \[\frac{I_{25}}{I_{0}} = \frac{\sin^2(15.41)}{51^2 \sin^2(\frac{15.41}{51})}\] Calculate the intensity ratio: \[\frac{I_{25}}{I_{0}} \approx 0.00123\] The peak intensity of the twenty-fifth secondary maximum is approximately 0.00123 times the intensity of the primary maximum. #Summary#: 1. The angular position of the twenty-fifth secondary maximum is approximately \(1.733^\circ\). 2. The peak intensity of the twenty-fifth secondary maximum is approximately 0.00123 times the intensity of the primary maximum.