The distance between the first and fifth minima of a single-slit diffraction pattern is 0.500 mm with the screen 37.0 cm away from the slit, when light of wavelength 530 nm is used. find the slit width.
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]D=37.0 cm=0.37 m[/tex] [tex]\lambda=530 nm=5.3 \cdot 10^{-7} m[/tex] while the distance between the first and the fifth minima is [tex]y_5-y_1 = 0.500 mm=0.5 \cdot 10^{-3} m[/tex] (2)
If we use the formula to rewrite [tex]y_5, y_1[/tex], eq.(2) becomes [tex] \frac{5 \lambda D}{a} - \frac{1 \lambda D}{a} =\frac{4 \lambda D}{a}= 0.5 \cdot 10^{-3} m [/tex] Which we can solve to find a, the width of the slit: [tex]a= \frac{4 \lambda D}{0.5 \cdot 10^{-3} m}= \frac{4 (5.3 \cdot 10^{-7} m)(0.37 m)}{0.5 \cdot 10^{-3} m}= 1.57 \cdot 10^{-3} m=1.57 mm[/tex]
