Respuesta :
Answer:
0.629 m
[tex]23.22^{\circ}, 36.26^{\circ}, 52.05^{\circ}, 80.3^{\circ}[/tex]
Explanation:
For destructive interference we have the case
[tex]sin\theta=\dfrac{m\lambda}{a}[/tex]
m = 1,2,3.....
Frequency is given by
[tex]f=\dfrac{75}{60}=1.25\ Hz[/tex]
Wavelength
[tex]\lambda=\dfrac{v}{f}\\\Rightarrow \lambda=\dfrac{15.5}{1.25}\\\Rightarrow \lambda=12.4\ cm[/tex]
The angle
[tex]\theta=tan^{-1}\dfrac{0.623}{3.1}\\\Rightarrow \theta=11.36^{\circ}[/tex]
The width is
[tex]a=\dfrac{\lambda}{sin\theta}\\\Rightarrow a=\dfrac{0.124}{sin11.36}\\\Rightarrow a=0.629\ m[/tex]
The width of the hole is 0.629 m
For destructive interference
[tex]sin\theta_2=\dfrac{2\lambda}{a}\\\Rightarrow \theta_2=sin^{-1}\dfrac{2\times 0.124}{0.629}\\\Rightarrow \theta_2=23.22^{\circ}[/tex]
[tex]\theta_3=sin^{-1}\dfrac{3\times 0.124}{0.629}\\\Rightarrow \theta_3=36.26^{\circ}[/tex]
[tex]\theta_4=sin^{-1}\dfrac{4\times 0.124}{0.629}\\\Rightarrow \theta_4=52.05^{\circ}[/tex]
[tex]\theta_5=sin^{-1}\dfrac{5\times 0.124}{0.629}\\\Rightarrow \theta_5=80.3^{\circ}[/tex]
The angles are [tex]23.22^{\circ}, 36.26^{\circ}, 52.05^{\circ}, 80.3^{\circ}[/tex]
Diffraction allows finding the answers for the size of the hole in the barrier and the angles where there are no waves are:
A) Size of the hole is d = 0.629 m
B) angles of destructive diffraction are: 23.2º, 36.3º, 52.1º and 80.3º
The diffraction phenomenon occurs when a coherent wave reaches an obstacle that has the size of the order of magnitude of the wavelength, there are two possibilities:
- Constructive. It occurs when the waves that come out are added together resulting in a great waves.
- Destructive. When the waves that come out have a path difference and the resulting wave has zero amplitude , which is expressed by the equation
d sin θ = m λ
m = 1, 2, 3, ...
Where d is the hole size, θ the angle, λ the wavelength and m is an integer.
They indicate that 75 waves pass every minute, let's find the frequency.
f = [tex]\frac{number_{wave}}{t}[/tex]
f = [tex]\frac{75}{60}[/tex]
f = 1.25 Hz
The speed of a wave is related to the product of its wavelength and its frequency, the speed of the wave is indicated v = 15.5 cm / s = 0.15 m/s.
v = λ f
λ = [tex]\frac{v}{f}[/tex]
λ = [tex]\frac{0.155}{1.25}[/tex]
λ = 0.124 m
Let's use trigonometry to find the angle of the destructive diffraction, indicate the distance to the shore x = 3.1m and the perpendicular distance is y = 62.3 cm = 0.623 m.
tan θ = [tex]\frac{y}{x}[/tex]
θ = tan⁻¹ [tex]\frac{y}{x}[/tex]
θ = tan⁻¹ [tex]\frac{0.623}{3.10}[/tex]
θ = 11.36º
Now we can substitute in the destructive diffraction expression, for the first zero (m = 1)
d = [tex]\frac{m \lambda }{sin \theta}[/tex]
d = [tex]\frac{1 \ 0.124}{sin \ 11.36}[/tex]
d = 0.629 m
This is the size of the opening in the barrier.
Part B. For other angles there are no waves.
To find the other points of destructive diffraction we change the integer.
[tex]sin \theta = \frac{m \ \lambda}{d}[/tex]
[tex]\theta = sin ^{-1} \frac{m \lambda}{d}[/tex]
m = 2
[tex]\theta_2 = sin ^{-1} \frac{2 \ 0.124}{0.629y}[/tex]
θ₂ = 23.2º
m = 3
[tex]\theta_3= sin^{-1} \frac{3 \ 0.124}{0.629}[/tex]
θ₃ = 36.3º
m = 4
[tex]\theta_4 = sin^{-1} \frac{4 \ 0.124 }{0.629}[/tex]
θ₄ = 52.1º
m = 5
[tex]\theta_5 = sin^{-1} \frac{5 \ 0.124 }{0.629}[/tex]
θ₅ = 80.3º
In conclusion using the expression for destructive diffraction we can find the answers for the size of the hole in the barrier and the angles where there are no waves are:
A) Size of the hole is d = 0.629 m
B) Angles of destructive diffraction are: 23.2º, 36.3º, 52.1º and 80.3º
Learn more here: brainly.com/question/23467404
