To solve this problem we will apply the concepts related to the double slit-experiment. Under this concept we understand the relationship between the minimum angle, depending on the order of the fringes, the wavelength and the distance between slits. Therefore we have the following relation,
[tex]sin(\theta_{min}) = \frac{m\lambda}{D}[/tex]
Here,
m = Order of the fringes
D = Distance between slits
[tex]\lambda[/tex] = Wavelength
Replacing with our values we have,
[tex]sin(\theta_{min}) = \frac{(1)(737*10^{-9})}{71.7*10^{-6}m}[/tex]
[tex]sin(\theta_{min}) = 0.01028[/tex]
Through the relationship between distances then we have that the basic amplification distance is given by the relationship between the distance of the slit L and the angle, then
[tex]Y=Lsin\theta[/tex]
[tex]Y =2.27*0.01028[/tex]
[tex]Y =0.0233m[/tex]
Thus the width of the central maximum is
[tex]W=2Y=2*0.0233[/tex]
[tex]W = 0.0466m[/tex]
Therefore the widht is 0.466m
The width, in centimeters, of the pattern's central maximum is; w = 46.67 cm
We are given;
Width of slit; d = 71.7 μm = 71.7 × 10^(-6) m
Wavelength of light; λ = 737 nm = 737 × 10^(-9) m
Distance from screen; y = 2.27 m
Now, we know that;
λ = d sin θ
Where θ is angle at first minimum
Thus;
sin θ = λ/d
sin θ = (737 × 10^(-9))/(71.7 × 10^(-6))
sin θ = 0.0102789
Now, the formula for the width of the pattern's central maximum is;
w = 2y sin θ
Thus;
w = 2 × 2.27 × 0.0102789
w = 0.04667 m
Converting to cm gives;
w = 46.67 cm
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