Question
Light with λ = 530 nm passes through a single slit and then illuminates a screen 9.0 m away. If the distance on the screen from the first dark fringe to the center of the interference pattern is 3.20 mm, what is the width of the slit?
Answers
tanα=x/L=3.2•10⁻³/9=3.56•10⁻⁴
tanα≈sinα
bsinα=kλ
b= kλ/sinα=1•530•10⁻⁹/3.56•10⁻⁴=1.5•10⁻³ m
tanα≈sinα
bsinα=kλ
b= kλ/sinα=1•530•10⁻⁹/3.56•10⁻⁴=1.5•10⁻³ m
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