1. Light with wavelength of 500 nm passes through a slit 0.2 mm wide. The diffraction pattern on a screen 60 cm away. Determine the distance between the central maximum and the second minimum.

__Known :__

λ = 500 nm = 500 x 10^{-9 }m = 5 x 10^{-7} m

d = 0.2 mm = 0.2 x 10^{-3} m = 2 x 10^{-4} m

l = 60 cm = 0.6 m

n = 2

__Wanted ____:__ y ?

__Solution :__

The width of the slit is minimal compared to the distance between the slit and the screen so that the angle is minimal (the width of the slit in the figure above is enlarged). The angle is so small that the sin θ ≈ tan θ.

sin θ ≈ tan θ = y / l = y / 0.6

Equation of diffraction by a single slit (minima) :

d sin θ = n λ

(2 x 10^{-4})(y/0,6) = (2)(5 x 10^{-7})

(2 x 10^{-4}) y = (0.6)(10 x 10^{-7})

(2 x 10^{-4}) y = 6 x 10^{-7}

y = (6 x 10^{-7}) / (2 x 10^{-4})

y = 3 x 10^{-3}

y = 0.003 m

y = 3 mm

2. Monochromatic light with wavelength of 5000 Å (1 Å = 10^{−10} m) passes through the single slit, produces diffraction pattern the first maximum as shown in figure. Determine the wide of slit.

__Known :__

λ = 5000 Å = 5000 x 10^{-10} m = 5 x 10^{-7} m

sin 30^{o }= 0,5

n = 1

__Wanted :__ wide of slit (d) ?

__Solution :__

d sin θ = n λ

d (0.5) = (1)(5 x 10^{-7})

d = (5 x 10^{-7}) / (0.5)

d = 10 x 10^{-7 }m

d = 1 x 10^{-6} m

d = 1 x 10^{-3} mm

d = 0.001 mm