Light of wavelength 520 nm passes through a slit of width 0.220 mm. (a) the width of the central maximum on a screen is 8.30 mm. how far is the screen from the slit?
In the single-slit experiment, the displacement of the minima of the diffraction pattern on the screen is given by [tex]y_n= \frac{n \lambda D}{a}[/tex] (1) where n is the order of the minimum y is the displacement of the nth-minimum from the center of the diffraction pattern [tex]\lambda[/tex] is the light's wavelength D is the distance of the screen from the slit a is the width of the slit
In our problem, [tex]\lambda=520 nm=5.2 \cdot 10^{-7} m[/tex] [tex]a=0.22 mm=0.22 \cdot 10^{-3} m[/tex] while the width of the central maximum on the screen corresponds to twice the distance of the first minimum from the center, and it is equal to [tex]2 y_1 = 8.30 mm=8.3 \cdot 10^{-3} m[/tex] Therefore the distance of the first minimum from the center is [tex]y_1 = \frac{8.3 \cdot 10^{-3} m}{2}=4.15 \cdot 10^{-3} m[/tex]
If we plug these numbers into eq.(1), we can find D, the distance of the screen from the slit: [tex]D= \frac{y_1 a}{ 1 \lambda }= \frac{(4.15 \cdot 10^{-3} m)(0.22 \cdot 10^{-3} m)}{(1)(5.2 \cdot 10^{-7} m)}= 1.76 m[/tex]
