Answer:
The width of the slit (in mm) is [tex]1.178*10^{-1}\ mm[/tex]
Explanation:
For Dark Band in diffraction:
dsinΘ=mλ...................Eq(1)
where:
d is the width of slit
If sinΘ is very small,then
Θ=x/D
where:
x is the distance of band from center of screen
D is the distance of screen from slit
Eq(1) will become:
[tex]d\frac{x}{D}=m\lambda[/tex]
For First dark fringe, m=1
[tex]d\frac{x_1}{D}=1*\lambda[/tex]..............Eq(2)
For third Dark fringe,m=3
[tex]d\frac{x_3}{D}=3\lambda[/tex]..................Eq(3)
Eq(3)-Eq(2)
[tex]d\frac{x_3}{D}-d\frac{x_1}{D}=3 \lambda -\lambda\\\frac{d}{D}(x_3-x_1)=\lambda(3-1)\\d=\frac{2*\lambda*d}{x_3-x_1}[/tex]
Where:
[tex]x_3-x_1=7.5*10^{-3} m\\D=75*10^{-2} m\\\lambda=589*10^{-9} m[/tex]
Now:
[tex]d=\frac{2*589*10^{-9}*75*10^{-2}}{7.5*10^{-3}}\\d=1.178*10^{-4}\ m[/tex]
In mm, it is given by:
[tex]d=1.178*10^{-4}*10^3\ mm\\d=1.178*10^{-1}\ mm[/tex]
The width of the slit (in mm) is [tex]1.178*10^{-1}\ mm[/tex]
We have that the the width of the slit is mathematically given as
d = 1.178×10^{−4}m=0.1178mm
From the question we are told
Generally the equation for the width of fringe is mathematically given as
x= (D/d) [(2(1) – 1)λ/2]
x= (D/d) [(1) λ/2]
x' = (D/d) [(2(3) – 1)λ/2]
x'= (D/d) [(5)λ/2]
We have
d = (75 x 10-2/ 7.5 x 10-3) [(4) (589 x 10-9)/2]
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