Answer:
[tex] w = \frac{2(590x10^{-9}) (1.5m)}{0.45x10^{-3}}=3.93 x10^{-3} m[/tex]
And if we convert this into mm we got:
[tex] w= 3.93 x10^{-3} m *\frac{1000mm}{1 m}= 3.93 mm[/tex]
Step-by-step explanation:
For this case we have the following info givenL
[tex] \lambda = 590 nm = 590 x10^{-9}m [/tex] represent the wavelength
[tex] s = 0.45 mm *\frac{1m}{1000 mm}= 0.45x10^{-3} m[/tex] represent the width of the single slit
[tex] L = 1.5 m[/tex] represent the longitude
And we want to find the maximum width of the central maximum in mm, so we can use the following formula:
[tex] w s = 2 \lambda L[/tex]
And if we solve for w the width we got:
[tex] w = \frac{2\lambda L}{s}[/tex]
And replacing we got:
[tex] w = \frac{2(590x10^{-9}) (1.5m)}{0.45x10^{-3}}=3.93 x10^{-3} m[/tex]
And if we convert this into mm we got:
[tex] w= 3.93 x10^{-3} m *\frac{1000mm}{1 m}= 3.93 mm[/tex]
