Question

A single slit 1.0 mm wide is illuminated by 450-nm light. What is the width of the central maximum (in cm) in the diffraction pattern on a screen 5.0 m away?

Answer 1

Width of the slit \(=1.0 \mathrm{~mm}=1.0 \times 10^{-3} \mathrm{~m}\)

Wavelength of the light \(\lambda=450 \mathrm{~nm}=450 \times 10^{9} \mathrm{~m}\)

The distance between the screen and slit is, \(D=5.0 \mathrm{~m}\)

The distance from the center to the first maximum is, \(x=\frac{\lambda D}{d}=\frac{\left(450 \times 10^{-9} \mathrm{~m}\right)(5.0 \mathrm{~m})}{1.0 \times 10^{-3} \mathrm{~m}}=2250 \times 10^{-6} \mathrm{~m}\)

The width of the central maximum is the twice of the distance from the center to the first maximum. Therefore, \(\beta=2 x=2\left(2250 \times 10^{-6} \mathrm{~m}\right)=4500 \times 10^{-6} \mathrm{~m}=0.4500 \mathrm{~cm}\)