\begin{tabular}{l} A diffraction grating is 3 cm long and has 3000 slits. White light shines through it and \\ creates fringes on a screen 0.7 m away. The first fringe of the color pewter appears \\ 3.4 cm from the central white fringe. What wavelength is this pewter light (in \\ nanometers, nm) \\ Question 14 \\ Saffron light has a wavelength of 440 nm. it shines through a single slit 0.4 mm wide. \\ If a screen is 2.5 m away, how WIDE is the central fringe (from lower dark fringe to \\ higher dark fringe)? \\ Q \\ \hline pts \\ \hline\end{tabular}

To determine the wavelength of the pewter light, we can use the equation for diffraction maxima: \[ d \sin(\theta) = n \lambda \] where \(d\) is the distance between slits, \(n\) is the order of the fringe (for the first fringe, \(n = 1\)), and \(\lambda\) is the wavelength. Given that the grating is 3 cm long with 3000 slits, the distance between slits \(d\) is: \[ d = \frac{3 \, \text{cm}}{3000} = 0.001 \, \text{cm} = 10^{-5} \, \text{m} \] The distance \(x\) from the central fringe to the pewter fringe is 3.4 cm, which is 0.034 m, and the distance to the screen \(L\) is 0.7 m. We can find \(\theta\) using: \[ \tan(\theta) = \frac{x}{L} = \frac{0.034}{0.7} \] For small angles, \(\sin(\theta) \approx \tan(\theta)\): \[ \sin(\theta) \approx \frac{0.034}{0.7} \approx 0.04857 \] Now plugging this back into the equation: \[ 10^{-5} \times 0.04857 = 1 \times \lambda \implies \lambda = 4.857 \times 10^{-7} \, \text{m} \] Converting to nanometers: \[ \lambda = 485.7 \, \text{nm} \] For the saffron light shine through a single slit, we can calculate the width of the central fringe using the equation for single-slit diffraction: \[ w = \frac{2L\lambda}{a} \] where \(a\) is the width of the slit. Here, \(L = 2.5 \, \text{m}\), \(\lambda = 440 \times 10^{-9} \, \text{m}\), and \(a = 0.4 \, \text{mm} = 0.4 \times 10^{-3} \, \text{m}\): \[ w = \frac{2 \cdot 2.5 \cdot 440 \times 10^{-9}}{0.4 \times 10^{-3}} \] Calculating that gives: \[ w \approx \frac{2 \cdot 2.5 \cdot 440 \times 10^{-9}}{0.4 \times 10^{-3}} \approx 0.006875 \, \text{m} = 6.875 \, \text{cm} \] So the width of the central fringe is approximately 6.875 cm.