Diffraction by a single Slit The central bright fringe in a single-slit diffraction pattern from light of wavelength 476 nm is \(2.0 \mathrm{cm}\) wide on a screen that is $1.05 \mathrm{m}$ from the slit. (a) How wide is the slit? (b) How wide are the first two bright fringes on either side of the central bright fringe? (Define the width of a bright fringe as the linear distance from minimum to minimum.)

The central bright fringe is formed at the maximum angle of diffraction. We can determine the angle using the formula for the linear width of the central bright fringe: \(W_c = 2 L \cdot \tan{\frac{\Theta_c}{2}}\) Where \(W_c\) is the width of the central bright fringe, \(L\) is the distance between the screen and the slit, and \(\Theta_c\) is the angle formed by the central bright fringe. We can solve for \(\Theta_c\) by rearranging the formula: \(\Theta_c = 2 \arctan{\frac{W_c}{2L}}\) Using the information given: \(\Theta_c = 2 \arctan{\frac{2.0 \times 10^{-2} m}{2 \times 1.05 m}} = 0.0191\) radians

The formula for single-slit diffraction is given by: \(a \sin{\Theta_c} = m\lambda\) Where \(a\) is the width of the slit, \(\Theta_c\) is the angle calculated in step 1, \(m\) is the order of the fringe (1 for the central bright fringe), and \(\lambda\) is the wavelength of the light. We can rearrange the formula to solve for the width of the slit: \(a = \frac{m\lambda}{\sin{\Theta_c}}\) Using the information given: \(a = \frac{1 \times 476 \times 10^{-9} m}{\sin{0.0191}} = 2.52 \times 10^{-5} m\) The width of the slit is \(2.52 \times 10^{-5} m\) or \(25.2 \mu m\).

The width of the first two bright fringes can be calculated using the same formula as in step 1, but we need to find the angles for the second and third bright fringes (\(\Theta_2\) and \(\Theta_3\)). Using the diffraction formula for multiple orders: \(a \sin{\Theta_2} = 2 \lambda\) \(a \sin{\Theta_3} = 3 \lambda\) Now, we can solve for the angles: \(\Theta_2 = \arcsin{\frac{2 \lambda}{a}} = 0.0378\) radians \(\Theta_3 = \arcsin{\frac{3 \lambda}{a}} = 0.0579\) radians Using these angles, we can calculate the linear width of the first two bright fringes: \(W_2 = 2L \cdot \tan{\frac{\Theta_2}{2}}\) \(W_3 = 2L \cdot \tan{\frac{\Theta_3}{2}}\) \(W_2 = 2 \times 1.05 m \cdot \tan{\frac{0.0378}{2}} = 0.0400 m\) \(W_3 = 2 \times 1.05 m \cdot \tan{\frac{0.0579}{2}} = 0.0626 m\) So, the first two bright fringes on either side of the central bright fringe have widths of \(4.0 cm\) and \(6.26 cm\).