Final answer:
The width of the central principal maximum for eight slits illuminated by monochromatic light is found using the multiple-slit interference formula. After calculation, it is determined to be approximately 0.596 mm, corresponding to option (d).
Step-by-step explanation:
The width of the central principal maximum on a screen can be calculated using the formula for the angular width of the main maximum in multiple-slit interference, \( \theta = \dfrac{m \lambda}{d} \), where \( m = \pm1 \) for the first minimum on either side of the central maximum, \( \lambda \) is the wavelength of the light, and \( d \) is the slit separation.
For eight slits equally separated by 0.149 mm (or \( 0.149 \times 10^{-3} \) meters) and light at \( \lambda = 523 \) nm (or \( 523 \times 10^{-9} \) meters), the angle \( \theta \) can be found using the formula \( \theta = \dfrac{\lambda}{d} \). Then, using the small angle approximation, the width of the central maxima (\( w \)) on the screen 2.35 m away can be calculated with \( w = 2 \times \theta \times L \), where \( L \) is the distance to the screen.
Plugging the given numbers into the formula, the width of the central principal maximum is found to be approximately 0.596 mm, which corresponds to option (d).