Respuesta :
Answer:
[tex]8.8\times 10^{-6} W/m^2[/tex]
Explanation:
We are given that
Wavelength,[tex]\lambda=571 nm=571\times 10^{-9} m[/tex]
[tex]1 nm=10^{-9} m[/tex]
R=75 cm=[tex]\frac{75}{100}=0.75 m[/tex]
1 m=100 cm
[tex] d=0.640 mm=0.64\times 10^{-3} m[/tex]
[tex]1 mm=10^{-3} m[/tex]
[tex]a=0.434 mm=0.434\times 10^{-3} m[/tex]
[tex]y=0.830 mm=0.830\times 10^{-3} m[/tex]
[tex]I_0=5.00\times 10^{-4}W/m^2[/tex]
[tex]tan\theta=\frac{y}{R}[/tex]
[tex]\theta=\frac{0.830\times 10^{-3}}{0.75\times 10^{-2}}[/tex]
[tex]\theta=1.1\times 10^{-3}rad[/tex]
[tex]tan\theta\approx \theta[/tex]
Because [tex]\theta[/tex] is small.
[tex]\phi=\frac{2\pi dsin\theta}{\lambda}[/tex]
[tex]\sin\theta\approx \theta[/tex],
Therefore
[tex]\phi=\frac{2\times\pi\times 0.64\times 10^{-3}\times 1.1\times 10^{-3}}{571\times 10^{-9}}[/tex]
[tex]\phi=7.74 rad[/tex]
[tex]\beta=\frac{2\pi a\theta}{\lambda}[/tex]
[tex]\beta=5.3 rad[/tex]
[tex]I=I_0cos^2(\frac{\phi}{2})(\frac{sin\frac{\beta}{2}}{\frac{\beta}{2}})^2[/tex]
[tex]I=5\times 10^{-4}cos^2(\frac{7.74}{2})(\frac{sin\frac{5.3}{2}}{\frac{5.3}{2}})^2[/tex]
[tex]I=8.8\times 10^{-6} W/m^2[/tex]
The intensity at a point on the screen that is 0.830mm from the center of the central maximum is; I = 8.64 × 10⁻⁶ W/m²
We are given;
Wavelength; λ = 571 nm = 571 × 10⁻⁹ m
Distance of interference from screen; R = 75 cm = 0.75 m
center to center distance between slits; d = 0.640 mm = 0.00064 m
width of each slit; a = 0.434 mm = 0.000434 m
Intensity at center; I₀ = 5 × 10⁻⁴ W/m²
Distance of intensity from center of the central maximum; y = 0.83 mm = 0.00083 m
Let us first find the angle θ from the formula;
tan θ = y/R
tan θ = 0.00083/0.75
tan θ = 0.001107
θ = tan⁻¹0.001107
θ = 0.06343
θ is very small and as such we use the formula;
∅ = (2πd sinθ)/λ
∅ = (2π * 0.00064 * sin 0.06343)/(571 × 10⁻⁹)
∅ = 7.7964 rad
Similarly, for distance between the centers of slit;
β = (2πa sinθ)/λ
β = (2π * 0.000434 * sin 0.06343)/(571 × 10⁻⁹)
β = 5.287 rad
Formula for the Intensity at a point on the screen that is 0.830mm from the center of the central maximum is;
[tex]I = I_{o}(cos^{2}\frac{\phi}{2}) [\frac{sin\frac{\beta}{2}}{\frac{\beta}{2}}]^{2}[/tex]
[tex]I = 5 * 10^{-4} }(cos^{2}(\frac{7.7964}{2})) [\frac{sin\frac{5.287}{2}}{\frac{5.287}{2}}]^{2}[/tex]
I = 8.64 × 10⁻⁶ W/m²
Read more about Intensity at; https://brainly.com/question/18190891
