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1 year ago

what principle is responsible for alternating light and dark bands when light passes through two or more narrow slits?

1 answer:

7 0

The superposition principle is responsible for alternating light and dark bands when light passes through two or more narrow slits.

The intensity pattern that appears on the lit screen is determined by the superposition principle. When the difference in pathways from the two slits to a location on the screen equals an integral number of wavelengths (0,λ,2λ ,...), constructive interference takes place.

The fact that the two waves' crests follow different paths ensures that they do. A distinctive pattern of brilliant and dark fringes is seen when monochromatic light illuminates a distant screen after passing through two small openings. The superposition of overlapping light waves coming from the two slits results in this interference pattern.

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what is the density of a substance that has a mass of 2.0 g , and when placed in a graduated cylinder the volume changed from 70

lubasha [3.4K]

A material with a mass of 2.0 g when placed in a graduated cylinder the volume changed from 70 ml to 75 ml has a density of 0.4 g/mL.

How do I calculate the substance's density?

We'll start by getting the substance's volume. This is attainable as follows:

Water volume: 70 mL

75 mL = volume of material + water.

Substance volume =?

Substance volume equals (substance volume plus water) - (Volume of water)

Substance volume = 75 - 70

5 mL is the substance's volume.

Finally, we will calculate the substance's density. Below is an example to help:

2.0 g is the substance's mass.

5 mL is the substance's volume.

Substance density =?

Mass / volume equals density.

Substance density = 2/5

0.4 g/mL is the substance's density.

The density is therefore 0.4 g/mL.

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6 0

1 year ago

A mechanic pushes a 2.30 ✕ 103-kg car from rest to a speed of v, doing 4,800 J of work in the process. During this time, the car

Illusion [34]

The horizontal force applied is 160 N while the velocity is 2.03 m/s.

<h3>What is the speed of the car?</h3>

The work done by the car is obtained as the product of the force and the distance;

W = F x

F = ?

x = 30.0 m

W = 4,800 J

F = 4,800 J/30.0 m

F = 160 N

But F = ma

a = F/m

a = 160 N/2.30 ✕ 10^3-kg

a= 0.069 m/s

Now;

v^2 = u^2 + 2as

u = 0/ms because the car started from rest

v = √2as

v = √2 * 0.069 * 30

v = 2.03 m/s

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5 0

2 years ago

In si units, the electric field in an electromagnetic wave is described by ey = 104 sin(1.40 107x − ωt). (a) find the amplitude

melamori03 [73]

Answers:
(a) $B_o = 0.3466$ μT
(b) $\lambda = 0.4488$ μm
(c) f = $6.68 * 10^{14}Hz$

Explanation:
Given electric field(in y direction) equation:
$E_y = 104sin(1.40 * 10^7 x -\omega t)$

(a) The amplitude of electric field is $E_o = 104$ . Hence

The amplitude of magnetic field oscillations is $B_o = \frac{E_o}{c}$
Where c = speed of light

Therefore,
$B_o = \frac{104}{3*10^8} = 0.3466$ μT (Where T is in seconds--signifies the oscillations)

(b) To find the wavelength use:
$\frac{2 \pi }{\lambda} = 1.40 * 10^7$
$\lambda = \frac{2 \pi}{1.40} * 10^{-7}$
$\lambda = 0.4488$ μm

(c) Since c = fλ
=> f = c/λ

Now plug-in the values
f = (3*10^8)/(0.4488*10^-6)
f = $6.68 * 10^{14}Hz$

6 0

3 years ago

A 300 MHz electromagnetic wave in air (medium 1) is normally incident on the planar boundary of a lossless dielectric medium wit

Masja [62]

Answer:

Wavelength of the incident wave in air = 1 m

Wavelength of the incident wave in medium 2 = 0.33 m

Intrinsic impedance of media 1 = 377 ohms

Intrinsic impedance of media 2 = 125.68 ohms

Check the explanation section for a better understanding

Explanation:

a) Wavelength of the incident wave in air

The frequency of the electromagnetic wave in air, f = 300 MHz = 3 * 10⁸ Hz

Speed of light in air, c = 3 * 10⁸ Hz

Wavelength of the incident wave in air:

$\lambda_{air} = \frac{c}{f} \\\lambda_{air} = \frac{3 * 10^{8} }{3 * 10^{8}} \\\lambda_{air} = 1 m$

Wavelength of the incident wave in medium 2

The refractive index of air in the lossless dielectric medium:

$n = \sqrt{\epsilon_{r} } \\n = \sqrt{9 }\\n =3$

$\lambda_{2} = \frac{c}{nf}\\\lambda_{2} = \frac{3 * 10^{6} }{3 * 3 * 10^{6}}\\\lambda_{2} = 1/3\\\lambda_{2} = 0.33 m$

b) Intrinsic impedances of media 1 and media 2

The intrinsic impedance of media 1 is given as:

$n_1 = \sqrt{\frac{\mu_0}{\epsilon_{0} } }$

Permeability of free space, $\mu_{0} = 4 \pi * 10^{-7} H/m$

Permittivity for air, $\epsilon_{0} = 8.84 * 10^{-12} F/m$

$n_1 = \sqrt{\frac{4\pi * 10^{-7} }{8.84 * 10^{-12} } }$

$n_1 = 377 \Omega$

The intrinsic impedance of media 2 is given as:

$n_2 = \sqrt{\frac{\mu_r \mu_0}{\epsilon_r \epsilon_{0} } }$

Permeability of free space, $\mu_{0} = 4 \pi * 10^{-7} H/m$

Permittivity for air, $\epsilon_{0} = 8.84 * 10^{-12} F/m$

ϵr = 9

$n_2 = \sqrt{\frac{4\pi * 10^{-7} *1 }{8.84 * 10^{-12} *9 } }$

$n_2 = 125.68 \Omega$

c) The reflection coefficient,r and the transmission coefficient,t at the boundary.

Reflection coefficient, $r = \frac{n - n_{0} }{n + n_{0} }$

You didn't put the refractive index at the boundary in the question, you can substitute it into the formula above to find it.

$r = \frac{3 - n_{0} }{3 + n_{0} }$

Transmission coefficient at the boundary, t = r -1

d) The amplitude of the incident electric field is $E_{0} = 10 V/m$

Maximum amplitudes in the total field is given by:

$E = tE_{0}$ and $E = r E_{0}$

E = 10r, E = 10t

3 0

3 years ago

7. Two people are pushing a 40.0kg table across the floor. Person 1 pushes with a force of 490N

artcher [175]

Answer:

20.4 $m/s^{2}$

Explanation:

To start doing this problem, first draw a free body diagram of the table. My teacher always tells us to do this, and I find that it is very helpful. I have attached a free body diagram to this answer- take a look at it.

First, let us see if Net force = MA. To do that, we need to determine whether the object is at equilibrium horizontally. For an object to be at equilibrium, it either needs to be moving at a constant velocity or not moving at all. Also, if an object is at equilibrium, there will not be any acceleration. But we know that there IS acceleration horizontally, so it cannot be in equilibrium. If it is not in equilibrium, we can use the formula ∑F= ma.

Let us determine the net force. Since the object is moving horizontally, we can ignore the weight and normal force, because they are vertical forces. The only horizontal forces we need to worry about are the applied force and force of friction.

Applied force = 1055 N (490 + 565)

Friction force= Unknown

To find the friction force, use the kinetic friction formula, Friction = μkN

μk is the coefficient, which the problem includes- it is 0.613.

N is the normal force, which we have to find.

*To find the normal force, we have to determine if the object is at equilibrium VERTICALLY. Since it has no acceleration vertically (it's not moving up/down), it is at equilibrium. Now, when an object is at equilibrium in one direction, it means that all the forces in that direction are equal. What are our vertical forces? Weight (mg) and Normal force (N). So it means that the Normal force is equal to the Weight.

Weight = mg = (40)(9.8) = 392 N

Normal force = 392 N

Now, plug it back into the formula (μkN): (0.613)(392) = 240.296 N

Friction = 240.296 N

Now that we know the friction, we can find the horizontal net force. Just subtract the friction force, 240.296 from the applied force, 1055 N

Horizontal Net Force: 814.704 N

Now that we know the net force, plug in the numbers for the formula

∑F= ma.

814.704 = (40.0)(a)

*Divide on both sides)

a = 20.3676 m/s^2

Round it to 3 significant figures, to get:

20.4 $m/s^{2}$

7 0

3 years ago