Answer:
width of maximum intensity at the center of screen is 12 mm
Explanation:
Position of minimum intensity on the screen is given as
[tex]a sin\theta = N\lambda[/tex]
now we know that for first position of minimum intensity on the screen we have
[tex]N = 1[/tex]
now we know
[tex]a sin\theta = 1 \lambda[/tex]
[tex]a = 0.380 mm[/tex]
[tex]\lambda = 633 nm[/tex]
now we have
[tex](0.380 \times 10^{-3})sin\theta = (633 \times 10^{-9})[/tex]
[tex]\theta = 0.095 degree = 1.66 \times 10^{-3} rad[/tex]
now total angular width of central maximum is given as
[tex]\phi = 2\theta = 2(1.66 \times 10^{-3}) rad[/tex]
[tex]\phi = 3.33 \times 10^{-3} rad[/tex]
now linear width is given as
[tex]w = L\phi[/tex]
[tex]w = (3.55)(3.33 \times 10^{-3})[/tex]
[tex]w = 0.012 m = 12 mm[/tex]
