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A radar tower sends out a signal of wavelength Image for A radar tower sends out a signal of wavelength lambda. It is x meters tall, and it stands on the edge of the o. It is xmeters tall, and it stands on the edge of the ocean. A weatherballoon is released from a boat that is a distance d out to sea. The balloon floats up to an altitudeh.In this problem, assume that the boat and balloon are so far awayfrom the radar tower that the small angle approximationholds.a) Due to interference with reflections off the water, certainwavelengths will be weak when they reach the balloon. What is themaximum wavelength Image for A radar tower sends out a signal of wavelength lambda. It is x meters tall, and it stands on the edge of the o that will interfere destructively?b) What is the maximum wavelength Image for A radar tower sends out a signal of wavelength lambda. It is x meters tall, and it stands on the edge of the o that will interfere constructively?

Respuesta :

Manetho

Answer:

a)[tex]\lambda= \frac{2hx}{d}[/tex]

b)[tex]\lambda= \frac{4hx}{d}[/tex]

Explanation:

a)Regular destructive two-slit interference equations apply

therefore, we can write

[tex]dsin\theta = (m+0.5)\lambda[/tex]

[tex]x\frac{h}{d} = m\lambda+0.5\lambda[/tex]

[tex]\frac{hx}{d}=0.5\lambda[/tex]

[tex]\lambda= \frac{2hx}{d}[/tex]

b) Regular constructive two-slit interference equations apply

therefore, we can write

[tex]dsin\theta = m\lambda[/tex]

[tex]2x\frac{h}{d}=0.5\lambda[/tex]

[tex]\lambda= \frac{4hx}{d}[/tex]

Lanuel

a. The maximum wavelength that will interfere destructively is [tex]\lambda=\frac{2hx}{d}[/tex]

b. The maximum wavelength that will interfere constructively is [tex]\lambda=\frac{4hx}{d}[/tex]

Given the following data:

  • Wavelength = [tex]\lambda[/tex]
  • Distance, d = x meters.
  • m = 0

a. To determine the maximum wavelength that will interfere destructively, we would apply an interference experiment formula:

The interference experiment formula.

Mathematically, an interference experiment is given by this formula:

[tex]dsin \theta =m \lambda[/tex]

Where:

  • d is the distance between slits.
  • m is the order of fringe.
  • [tex]\lambda[/tex] is the wavelength.
  • [tex]\theta[/tex] is the angle between bright fringe.

For a destructive two-slit interference, we have:

[tex]dsin \theta =(m+ \frac{1}{2} ) \lambda[/tex]

Note: From the diagram the value of [tex]sin \theta[/tex] is [tex]\frac{h}{d}[/tex].

Substituting the given parameters into the formula, we have;

[tex]x\frac{h}{d} =(0+ \frac{1}{2} ) \lambda\\\\x\frac{h}{d} =\frac{1}{2} \lambda\\\\\lambda d = 2hx\\\\\lambda=\frac{2hx}{d}[/tex]

b. To determine the maximum wavelength that will interfere constructively:

[tex]2x\frac{h}{d} =(0+ \frac{1}{2} ) \lambda\\\\2x\frac{h}{d} =\frac{1}{2} \lambda\\\\\lambda d = 2hx\\\\\lambda=\frac{4hx}{d}[/tex]

Read more on wavelength here: https://brainly.com/question/14702686

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