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WARRIOR [948]
3 years ago
14

Explain how the aperture geometry relates to the diffraction pattern.

Physics
1 answer:
konstantin123 [22]3 years ago
5 0

Answer:

The answer to this question is given below in the explanation section.

Explanation:

how the aperture geometry relates to the

diffraction pattern:

Diffraction is the spreading out of waves as they pass through an aperture or around objects.it occurs when the size of the aperture or obstacle is of the same order of magnitude as the wavelength of  the incident wave.For every small aperture sizes,the vast majority of the wave is blocked.For large apertures the wave passes by or through the obstacle without any significant of diffraction.

            in an aperture with width smaller than the wavelength,the wave transmitted through the aperture spreads all the way around the behave like a point sources of waves.

                             single slit diffraction pattern

The diffraction pattern made by waves passing through a slit of width \alpha (larger than∫) can be understood by imagining a series of point sources all in phase along the width of the slit.The waves moving directly forward are all in phase,so they from a large central maximum.

                 if the waves travels at an angle Ф from the normal to the slit,then there is a path difference x between the waves production at the two end of the slit.

           x=a sinΦ

The path difference between the top and middle waves is λ/2 then they are exactly out of phase and cancel each other out. This happens to all consecutive pairs of waves (the ones produced by the second source from the top and the second source past the middle etc.)at the angle so there is no resultant wave at this angle.Thus a minimum is the diffraction pattern is obtained at

                                                   λ=α sinθ

Now slit can be divided into four equal sections and the pairing of sources to give destructive interference can be repeated for the top two section ,which is identical to the result of pairing off matching sources in the bottom two sections.in this case we obtained from the minimum.

             λ/2=α/4 sinθ

we can divided the slit aperture into six equal sections and pair off sources in the top two divisions and then the bottom two,to give destructive interference for every matched pair.The minimum of intensity are obtained at angles

                      nλ = α sinθ

where n is an integer (1,2,......),  but not n=0.There is a maximum of intensity in the center of the pattern. This process only gives the position of the minima,does not work for positions of the maxima,and so does not give the intensities of the maxima.

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Shadow is formed when an_____<br>object comes in the<br>way<br>of light.​
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shadow is formed when an opaque object comes on path of light shadows are formed because light travels in a straight line and it can't pass through an opaque object

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3 years ago
Read 2 more answers
4A. How high is a 12 kg monkey in a tree if it has 509 J of gravitational potential Energy?
True [87]

4A. PE = MxGxH. (You can consider g as 9.8 / 10m/s as well)

509 J = 12x10xH

509 J = 120xH

H = 509/120

H = 4.24 m

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3 years ago
A steel piano wire, of length 1.150 m and mass 4.80 g is stretched under a tension of 580.0 N.
kaheart [24]

A steel piano wire, of length 1.150 m and mass of 4.80 g is stretched under a tension of 580.0 N.the speed of transverse waves on the wire would be  372.77 m/s

<h3>What is a sound wave?</h3>

It is a particular variety of mechanical waves made up of the disruption brought on by the movements of the energy. In an elastic medium like the air, a sound wave travels through compression and rarefaction.

For calculating the wave velocity of the sound waves generated from the piano can be calculated by the formula

V= √F/μ

where v is the wave velocity of the wave travel on the string

F is the tension in the string of piano

μ is the mass per unit length of the string

As given in question a steel piano wire, of length 1.150 m and mass of 4.80 g is stretched under a tension of 580.0 N.

The μ is the mass per unit length of the string would be

μ = 4.80/(1.150×1000)

μ = 0.0041739 kg/m

By substituting the respective values of the tension on the string and the density(mass per unit length) in the above formula of the wave velocity

V= √F/μ

V=√(580/0.0041739)

V =  372.77 m/s

Thus,  the speed of transverse waves on the wire comes out to be  372.77 m/s

Learn more about sound waves from here

brainly.com/question/11797560

#SPJ1

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