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WARRIOR [948]
2 years ago
14

Explain how the aperture geometry relates to the diffraction pattern.

Physics
1 answer:
konstantin123 [22]2 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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A sound wave has a frequency of 425 hz. what is the period of this wave? 0.00235 seconds 0.807 seconds 425 seconds 850 seconds
77julia77 [94]

The period of the sound wave at the given frequency is determined as 0.00235 second.

<h3>Period of the sound wave</h3>

The period of the sound wave at the given frequency is calculated as follows;

Period is reciprocal of frequency.

T = 1/f

T = 1/425

T = 0.00235 second

Thus, the period of the sound wave at the given frequency is determined as 0.00235 second.

Learn more about period here: brainly.com/question/10428039

#SPJ1

3 0
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A virtual image is formed 17.0 cm from a concave mirror having a radius of curvature of 39.0 cm. (a) Find the position of the ob
Wittaler [7]

Explanation:

Image distance, v = -17 cm (-ve for virtual image)

Radius of curvature of concave mirror, R = 39 cm

Focal length, f = -19.5 cm (-ve for a concave mirror)

(a) Using mirror's formula as :

\dfrac{1}{v}+\dfrac{1}{u}=\dfrac{1}{f}

\dfrac{1}{u}=\dfrac{1}{f}-\dfrac{1}{v}

\dfrac{1}{u}=\dfrac{1}{-19.5}-\dfrac{1}{(-17)}  

u = 132.6 cm    

So, the object is placed 132.6 cm in front of the mirror.

(b) Magnification of the  mirror, m=\dfrac{-v}{u}

m=\dfrac{-17}{132.6}

m = -0.128

Hence, this is the required solution.

7 0
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A 0.0100-kilogram grape hangs from a vine 1.50 meters from the ground. What is the potential energy of the grape?
ludmilkaskok [199]
The potential energy is defined as Ep=m*g*h where m is the mass of the body, g=9.81 m/s² and h is the height of the body. In our case m=0.01 kg and h=1.5 m. So when we input the values into the equation:

Ep=0.01*9.81*1.5= 0.14715 J. 

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PLEASE HELP I NEED AN A FOR THIS TO PASS
Troyanec [42]

Answer:

B

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This is the correct answer because given that this is a closed system, this means that energy is not lost but only changes form. The fact that the ball bounces shorter each time shows that energy is being transformed and not lost. Because of this, the ball must be transforming kinetic energy into sound which is why B is correct

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