Final answer:
To find the total width of the central bright fringe in a two-slit pattern, one would typically need the wavelength and slit separation. However, given the fifth-order maxima are 80.0 cm apart, one can deduce that the total width of the central bright fringe is 0.32 m or 320 mm.
Step-by-step explanation:
The student has asked about the total width of the central bright fringe in a two-slit diffraction pattern. In answering this question, we must use the formula for the position of the m-th maximum in a double-slit interference pattern, which is given by:
y = (mλL)/d
where y is the distance from the central maximum to the m-th maximum on the screen, m is the order of the maximum, λ is the wavelength of light, L is the distance from the slits to the screen, and d is the separation between the slits.
To find the width of the central maximum, one would need to know the wavelength of the light and the separate between the slits. However, the question provides the distance between the fifth-order maxima, which are symmetric around the central maximum. Thus, the width of the central bright fringe is twice the distance from the central maximum to the first order maximum. Since the fifth maximum is at 80.0 cm or 0.80 m from the central maximum, and there are four maxima within this range, each will have a width of 0.80 m/5 = 0.16 m. Hence, the width of the central bright fringe is double that of the first maximum, which is 2 x 0.16 m = 0.32 m or 320 mm.