Answer:
The width is [tex]w_c = 0.00252 \ m[/tex]
Explanation:
From the question we are told that
The width of the single slit is [tex]a = 1.5 \ mm = 1.5 *10^{-3} \ m[/tex]
The wavelength is [tex]\lambda = 420 *10^{-9} \ m[/tex]
The distance of the screen is [tex]D = 4.5 \ m[/tex]
Generally the width of the central maximum is
[tex]w_c = 2 * y[/tex]
where y is the width of the first maxima which is mathematically represented as
[tex]y = \frac{\lambda * D}{a}[/tex]
=> [tex]y = \frac{ 420 *10^{-9} * 4.5}{ 1.5*10^{-3}}[/tex]
=> [tex]y = 0.00126 \ m[/tex]
So
[tex]w_c = 2 *0.00126[/tex]
[tex]w_c = 0.00252 \ m[/tex]
