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Tju [1.3M]
4 years ago
5

Explain how the diffraction of light shows that light behaves like a wave.

Physics
2 answers:
wolverine [178]4 years ago
7 0

Answer: Particles cannot bend around the edges of an obstacle, as light waves does.

Explanation:

Diffraction happens when a wave (mechanical or electromagnetic wave) meets an obstacle or a slit .When this occurs, the wave bends around the edges of the obstacle or passes through the opening of the slit that acts as an obstacle, forming multiple patterns with the shape of the aperture of the slit.

Note this phenomenon is a characteristic of waves behaviour and not particles, because <u>particles cannot bend around the edges of an obstacle</u>, as waves (light waves in this case) does.

Oduvanchick [21]4 years ago
5 0

<u>Diffraction of light behaves like wave: </u>

Light always bends when it is passed through edge or slit. The property of bending of waves is known as diffraction. The diffraction pattern created when light bends around edge or slit shows that diffraction of light has wave like properties.  The particle passing through edge or slit does not bend.

When the light travels through prism which is a diffracting medium, it is broken down into its constituents which shows wavelength of the light with the amplitude of its existence and it follows frequency and period. These are all the property of a wave and not a particle, hence light is being demonstrated as wave in diffraction.

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seropon [69]

Answer:

the lowest possible frequency of the emitted tone is 404.79 Hz

Explanation:

   Given the data in the question;

S₁ ←  5.50 m → L

↑

2.20 m

↓

S₂

We know that, the condition for destructive interference is;

Δr = ( 2m + \frac{1}{2} ) × λ

where m = 0, 1, 2, 3 .......

Path difference between the two sound waves from the two speakers is;

Δr = √( 5.50² + 2.20² ) - 5.50

Δr = 5.92368 - 5.50

Δr = 0.42368 m

v = f × λ

f = ( 2m + \frac{1}{2})v / Δr

m = 0, 1, 2, 3, ....

Now, for the lowest possible frequency, let m be 0

so

f = ( 0 + \frac{1}{2})v / Δr

f = \frac{1}{2}(v) / Δr

we know that speed of sound in air v = 343 m/s

so we substitute

f = \frac{1}{2}(343) / 0.42368

f = 171.5 / 0.42368

f = 404.79 Hz

Therefore, the lowest possible frequency of the emitted tone is 404.79 Hz

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3 years ago
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Ira Lisetskai [31]

Answer:

At the origin (x' = 0 m)

Explanation:

Note: From the question, when the cart travels towards the origin, the magnitude of its exact position reduces with time.

The formula of speed is given as

S = d/t................. Equation 1

Where S = speed of the cart, d = distance covered by the cart over a certain time. t = time taken to cover the distance.

make d the subject of the equation,

d = St ................. Equation 2

Given: S = 2.0 m/s, t = 3.0 s

Substitute into equation 2

d = 2(3)

d = 6 m.

From the above, the cart covered a distance of 6 m in 3 s.

The exact position of the cart = Initial position-distance covered

x' = x-d............ Equation 3

Where x' = exact position of the cart 3 s later, x = initial position of the cart, d = distance covered by the cart in 3.0 s.

Given: x = +6.0 m, d = 6 m.

Substitute into equation 3

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3 years ago
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Answer:

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Now, for the cost we use the following formula:

Cost = (Energy)(Unit\ Cost)\\Cost = (4\ KWh)(10\ fills/KWh)\\

<u>Cost = 40 fills</u>

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