Question

How wide is the central diffraction peak on a screen 2.10 m behind a 0.0328-mm-wide slit illuminated by 568-nm light?

Answer 1

Answer:

0.0364 m

Explanation:

$D$ = Screen distance = 2.10 m

$d$ = Slit width = 0.0328 mm = 0.0328 x 10⁻³ m

$\lambda$ = Wavelength of light = 568 x 10⁻⁹ m

$w$ = width of the central diffraction peak

Width of the central diffraction peak is given as

$w = \frac{D \lambda}{d} \\w = \frac{(2.10) (568\times10^{-9})}{0.0328\times10^{-3}}\\w = 0.0364 m$