Final answer:
To calculate the width of the central maximum in a single-slit diffraction pattern, one can use the single-slit diffraction formula for the angle of the first minimum and then double this position since the central maximum spans from the first minimum on one side to the first minimum on the other side. For a single slit 1.7 mm wide with 480 nm light and a screen 6.5 m away, the width of the central maximum is approximately 0.36712 cm.
Step-by-step explanation:
The subject of the question is the calculation of the width of the central maximum in a single-slit diffraction pattern, given a slit width, light wavelength, and the distance to the screen. To find the width of the central maximum on the screen, we can use the formula for the angle of the first minimum in single-slit diffraction: θ = λ/d, where λ is the wavelength of the light and d is the slit width. This angle is then related to the position of the first minimum on the screen by the small angle approximation, x = Lθ, where L is the distance to the screen and x is the position on the screen. Since the central maximum spans from the first minimum on one side to the first minimum on the other side, its total width (W) is 2x. Using the given values, λ = 480 nm (converted to meters), d = 1.7 mm (converted to meters), and L = 6.5 m, we can calculate θ and then x:
θ = (480 x 10-9 m) / (1.7 x 10-3 m) = 2.824 x 10-4 radians
x = Lθ = 6.5 m × 2.824 x 10-4 radians = 0.0018356 m or 1.8356 mm
Therefore, the width of the central maximum (W) is 2x = 2 × 1.8356 mm = 3.6712 mm or 0.36712 cm.