Diffraction Understanding a solution

In summary, the conversation discusses the relationship between the diameter of a hole and the power of a beam of light passing through it. The participants calculate that halving the diameter of the hole will decrease its area by a factor of 16, and therefore decrease the power through it by 1/16. However, there is confusion about how this conclusion is reached, and the conversation delves into the concept of area being proportional to the square of the diameter. The conversation also mentions that the assumption is made that the incident beam of light is circular in shape.
  • #1
Hebel
8
0

Homework Statement


(see attached problem)

Homework Equations


(see attached problem)

The Attempt at a Solution


So assuming the area of the hole is circular, halving its diameter, will decrease its area by a factor of 16. Therefore LI will be I/16? However in the solution (see attached solution) it says: 'Halving the diameter of the hole will decrease the power through it by 1/4'. How does one make this conclusion?
 

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  • #2
Hebel said:
'Halving the diameter of the hole will decrease the power through it by 1/4'. How does one make this conclusion?
Because ##P=IA##, therefore any change in the scaling of ##A## will result in the same amount of scaling in ##P##.
 
  • #3
Hebel said:
So assuming the area of the hole is circular, halving its diameter, will decrease its area by a factor of 16. Therefore LI will be I/16? However in the solution (see attached solution) it says: 'Halving the diameter of the hole will decrease the power through it by 1/4'. How does one make this conclusion?
Hebel said:
halving its diameter, will decrease its area by a factor of 16.
perhaps some error is there- Area is proportional to (d/2)^2 !

i think one should remember 'as given' that halving the diameter will decrease the area of hole by a factor of four - but at the same time the diffracted beam increases its width by a factor of 2 - so it leads to a further spread of the same power in four times the area (decrease by a factor of four )so one gets a net 1/16 of the power.
 
  • #4
Why does twice the width mean 4 times the area?
 
  • #5
Hebel said:
Why does twice the width mean 4 times the area?
##A_1=\frac{1}{4}\pi D_1^2##. Now if you set ##D_2=2D_1##, what is ##A_2=\frac{1}{4}\pi D_2^2## in terms of ##A_1##?
 
  • #6
So the new area is 4 times the old one I see, so we are just assuming that the area of the beam incident on the screen will be circular right?
 
  • #7
Hebel said:
so we are just assuming that the area of the beam incident on the screen will be circular right?
You are supposed to assume that the illumination is of plane wave from the left.
 

FAQ: Diffraction Understanding a solution

What is diffraction?

Diffraction is a phenomenon that occurs when waves encounter an obstacle or pass through an opening and bend around it, resulting in interference patterns. It is commonly observed in light, sound, and water waves.

How does diffraction help us understand solutions?

Diffraction can be used to analyze the properties of solutions by measuring the patterns created when light is passed through them. These patterns can reveal information about the size and distribution of particles in the solution, as well as the concentration and interactions between different substances.

What is the difference between diffraction and refraction?

Diffraction involves the bending of waves around an obstacle or opening, while refraction is the change in direction of waves as they pass through different mediums. Diffraction is dependent on the wavelength of the waves, while refraction is dependent on the properties of the medium.

How is diffraction used in scientific research?

Diffraction techniques, such as X-ray diffraction, are commonly used in scientific research to determine the structures of molecules and crystals. Diffraction can also be used to study the properties of materials, including their size, shape, and composition.

What are some real-life applications of diffraction?

Diffraction has many practical applications, including in the fields of medicine, engineering, and materials science. X-ray diffraction is used in medical imaging to create detailed images of bones and tissues, while diffraction gratings are used in optical devices such as spectrometers and cameras. Diffraction is also used in the manufacturing of products such as CDs and DVDs, which rely on diffraction patterns to store and read data.

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