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4 years ago

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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How lubricant reduce friction

Tamiku [17]

Lubricant is a type of oil. It smoothes surfaces. Meaning anything rubbing against the other object wont produce enough heat, which is what creates friction. Its an ideal tool to use for a lot of things.

I hope this helps :)

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3 years ago

A crate rests on the flatbed of a truck that is initially traveling at 15 m/s on a level road. The driver applies the brakes and

lina2011 [118]

Answer:0.3

Explanation:

Given

velocity of car=15 m/s

truck brought to halt in a distance of 38 m

We know

$v^2-u^2=2as$

Final velocity (v)=0

$0-(15)^2=2(a)(38)$

$a=\frac{-225}{76}$

$a=-2.96 m/s^2$ (deceleration)

Therefore minimum coefficient of friction \mu will be

$\mu \times g=a$

$\mu =\frac{a}{g}$

$\mu =\frac{2.96}{9.8}=0.302$

7 0

3 years ago

Man A (70kg) and Man B (90kg) are hanging motionless from a platform at rest. What is the tension TA in the top rope if the plat

Damm [24]

Answer:

The tension in the upper rope (top rope), T1 = 1,888 N

Explanation:

The Parameters that were given:

Mass A, M1 = 70kg

Mass B. M2 = 90kg

acceleration, a = 2 m/s2

Assume the rope doesn't have mass, acceleration due to gravity, g

= 9.8 m/s2

The tension, T in a platform = m (a + g)

Then the tension, T1 in the upper rope = m1 (a + g) + T2

Where T2 = Tension in the lower rope

First, we calculate T2

Since the platform accelerates upward the acceleration would be positive

T2 = m2 (a + g)

T2 = 90kg ( 2 m/s2 + 9.8 m/s2)

T2 = 1,062N

To calculate the tension T1,

T1 = m1 (a + g) + T2

= 70kg (2 m/s2 + 9.8 m/s2) + 1062N

T1 = 1,888 N

3 0

3 years ago

The atmospheric features of Neptune are easier to see than those of Uranus because A. Neptune has greater warmth and less haze.

Lelechka [254]

If I’m being honest I think it’s a or b

4 0

3 years ago

Tap on the photo. For each diagram, explain why the light behaves in the way that it does.

dem82 [27]

Answer:

Diagram 1, 3 and 4 can be explained with the phenomenon of refraction.

Refraction occurs when a ray of light crosses the interface between two mediums with different optical density: when this occurs, the ray of light is bent and its speed changes, according to Snell's law

$n_1 sin \theta_1 = n_2 sin \theta_2$

where $n_1,n_2$ are the refractive index of the 1st and 2nd medium

$\theta_1, \theta_2$ are the angle that the incident ray and the refracted ray makes with the normal to the interface

In diagram, 1, the ray of light arrives perpendicularly to the interface, so it is refracted through the medium but it doesn't change its direction (only its speed).

In diagram 3, the ray of light is refracted twice: at the 1st interface and at the 2nd interface. In the 1st case, it goes from a medium with lower refractive index to a medium with higher refractive index ( $n_1$ ), this means that $\theta_2$ , so the ray bends towards the normal. Vice-versa, in the 2nd case the ray goes from a medium with higher refractive index to a medium with lower refractive index ( $n_1>n_2$ ), so it bends away from the normal ( $\theta_2>\theta_1$ ).

In diagram 4, the ray of light is also refracted twice. The ray of light here acts exactly the same as in diagram 3, h

However, this time the 2nd interface is the opposite direction with respect to diagram 3, so in this case the ray of light at the 2nd interface bends in the opposite direction (still away from the normal).

Diagram 2 instead is an example of reflection, that occurs when a ray of light bounces off the interface between the two mediums, withouth entering the 2nd medium.

According to the law of reflection:

- The incoming ray, the reflected ray and the normal to the boundary are all in the same plane

- The angle of incidence is equal to the angle of reflection (both are measured relative to the normal to the boundary)

Therefore in this diagram, the ray of light hits the boundary at approx. 45 degrees from the normal, and then it is reflected back approximately at 45 degrees on the other side with respect to the normal.

3 0

4 years ago