Respuesta :
Answer:
a) 0.0130 m
b') w' = =6.46*10^{-3] m
Explanation:
given data:
\lambda of light = 633 nm
width of siit a =0.360 mm
distance from screen = 3.75 m
a) the first minima is located at
[tex]sin\theta = \frac{\lambda}{a}[/tex]
=[tex]= \frac{633 *10^{-9}}{.360*10^{-3}}[/tex]
[tex]\theta = 0.100[/tex]
[tex]y_1 = dtan\theta_1 = 3.75*tan(0.100) = 6.54 *10^{-3} m[/tex]
with of central fringe = 2y_1 = 2*6.54 *10^{-3} = 0.0130 m
b)
width of the first bright fringe on either side of the central one = [tex]w' = y_2 -y_1[/tex]
calculation for y_2
[tex]sin\theta = 2\frac{\lambda}{a}[/tex]
= [tex]= 2*\frac{633 *10^{-9}}{.360*10^{-3}}[/tex]
[tex]\theta = 2*0.100 = 0.200 [/tex]
[tex]y_2 = dtan\theta_1 = 3.75*tan(0.200) =0.0130 m[/tex]
[tex]w' = 0.0130 -6.54 *10^{-3}[/tex]
w' = =6.46*10^{-3] m
