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1 year ago

What is the phase difference in a Young's double slit experiment where the intensity is half that of the central maximum?

Physics

1 answer:

emmasim [6.3K]1 year ago

4 0

The phase difference is π/2.

<h3>What is phase difference?</h3>

The phase difference is defined as the difference in the phase angle of the two waves.

Let, the intensity of light at the central maximum in Young's double slit experiment with two equal intensity sources = I₀

The intensity of light at an phase difference α is given by:

I= (I₀/2)(1 +cos²α)

According to the question: we have to find the phase difference in a Young's double slit experiment where the intensity is half that of the central maximum. Let it be α. So,

(I₀/2)(1 +cos²α) = I₀/2

⇒(1 +cos²α) = 1

⇒ cos²α = 0

⇒ α = π/2.

Hence, the phase difference is π/2.

Learn more about phase difference here:

brainly.com/question/14594671

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An animal-rescue plane flying due east at 55 m/s drops a bale of hay from an altitude of 69 m . The acceleration due to gravity

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Answer:

The momentum of the bale the moment it strikes the ground is 1076.68 kg-m/s.

Explanation:

It is given that,

Velocity of an animal-rescue plane, $v_x=55\ m/s$

It drops a bale of hay from an altitude of 69 m, h = 69 m

The vertical velocity of plane is given by :

$v_y=\sqrt{2gh}$

$v_y=\sqrt{2\times 9.81\times 69} =36.79\ m/s$

Weight of the bale of hay, W = 192 N

If m is the mass, then weight is given by :

$m=\dfrac{W}{g}$

$m=\dfrac{192}{9.81}=19.57\ kg$

The resultant momentum of the bale the moment it strikes the ground is given by :

$p=m(v_x-v_y)$

$p=19.57\times 55i-19.57\times 36.79j$

$p=(1076.35i-719.98j)\ kg-m/s$

Magnitude of momentum,

$p=\sqrt{1076.35^{2}+719.98}$

p = 1076.68 kg-m/s

So, the momentum of the bale the moment it strikes the ground is 1076.68 kg-m/s. Hence, this is the required solution.

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4 years ago

A proton is released from rest in a uniform electric field of magnitude 50000 v/m directed along the positive x axis and undergo

Svet_ta [14]

For the law of conservation of energy, the amount of kinetic energy the proton gained in $d=0.5 m$ is equal to the amount of electric potential energy it losts covering the same distance.

The potential difference across which the proton travelled is given by
$\Delta V = E d$
where E is the electric field intensity. Replacing the numbers, we get
$\Delta V = (50000 V/m)(0.5 m)=25000 V$

The electric potential energy lost by the proton is given by
$\Delta U = q \Delta V$
where $q=1.6 \cdot 10^{-19}C$ is the charge of the proton. Therefore, this quantity is equal to
$\Delta U = (1.6 \cdot 10^{-19}C)(25000 V)=4 \cdot 10^{-15}J$

And based on what we said at the beginning, this electric potential energy lost by the proton is exactly equal to the amount of kinetic energy it gained:
$\Delta K = 4 \cdot 10^{-15}J$

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3 years ago

In the water cycle, water vapor evaporates into the atmosphere and forms clouds through condensation. What happens in the atmosp

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The correct answer for this problem is C.

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3 years ago

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Answer:

$3 \times {10}^{8} m {s}^{ - 1}$

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3 years ago

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a bullet of mass 4g when fired with a velocity of 50m/s can enter a wall upto a depth of 10cm how much will be the average resis

jok3333 [9.3K]

Mass, m = 4g = 0.004 kg
Velocity, = 50cm/s = 0.5m/s
Distance, 10cm = 0.1m
The wall would have to resist the energy acquired by the bullet.

Kenetic Energy of bullet = Resistance offered by the wall.

1/2 mv² = Resistance Force * Distance

(1/2) * 0.04 * 0.5 * 0.5 = F * 0.1

0.5 * 0.04 * 0.5 * 0.5 = F * 0.1

0.5 * 0.04 * 0.5 * 0.5/0.1 = F

0.05 = F

Therefore, Resistance offered by the wall = 0.05 N

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3 years ago