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Nadya [2.5K]
3 years ago
13

The distance between slits in a double-slit experiment is decreased by a factor of 2. If the distance between fringes is small c

ompared to the distance from the slits to the screen, the distance between adjacent fringes near the center of the interference pattern _______.
Physics
1 answer:
Elodia [21]3 years ago
4 0

Answer:

the distance between adjacent fringes is increased by a factor o 2

Explanation:

To find how the distance between fringes is modified you can use the following formula for the calculation of the distance between fringes:

\Delta y=\frac{\lambda D}{d}

D: distance to the screen

d: distance between slits

λ: wavelength of the light

if d is decreased by a factor of 2, that is d'=1/2d, you have:

\Delta y'=\frac{\lambda D}{d'}=\frac{\lambda D}{(1/2)d}=2\Delta y

hence, the distance between adjacent fringes is increased by a factor o 2

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The vitreous humor, a transparent, gelatinous fluid that fills most of the eyeball, has an index of refraction of 1.34. Visible
Leto [7]

Answer:

a)   298.5 nm ,  522.4 nm  and b)  radiation frequency does not change

Explanation:

When electromagnetic radiation reaches a medium with a different index of refraction, the medium vibrates the molecules, as if it were a resonance process, whereby the medium vibrates at the same frequency as the incident light.

On the other hand, when the light reaches another medium its average speed within the medium changes, it is now less than the speed of light in a vacuum (c) for this to happen as we saw that the frequency is constant there must be a change in the wavelength of the radiation that is characterized by the ratio

    λₙ = λ₀ / n

    λₙ = 400 nm    in the void

    λₙ = 400 / 1.34

    λₙ= 298.5 nm

   λ₀ = 700 nm

   λₙ = 700 / 1.34

   λₙ = 522.4 nm

The radiation frequency does not change

4 0
3 years ago
Ultraviolet light from a distant star is traveling at 3.0 × 108 m/s. how long will it take for the light to reach earth if it mu
castortr0y [4]

Given speed and the distance that must be covered, the time it will take the ultraviolet light to reach the earth is 3.7 × 10⁴ hours.

<h3>What is Speed?</h3>

Speed is simply referred to as distance traveled per unit time.

Mathematically, Speed = Distance ÷ time.

Given the data in the question;

  • Speed of the Ultraviolet light c = 3.0 × 10⁸m/s = 1.08 × 10⁹km/h
  • Distance it must cover d = 4.0 × 10¹³km
  • Time elapsed t = ?

We substitute our given values into the expression above.

Speed = Distance ÷ time

1.08 × 10⁹km/h = 4.0 × 10¹³km ÷ t

t = 4.0 × 10¹³km ÷ 1.08 × 10⁹km/h

t = 3.7 × 10⁴ hrs

Therefore, given speed and the distance that must be covered, the time it will take the ultraviolet light to reach the earth is 3.7 × 10⁴ hours.

Learn more about speed here: brainly.com/question/7359669

#SPJ4

5 0
1 year ago
A -5.0 μC charge experiences a 11 i^ N electric force in a certain electric field. [Recall that i^ is a unit vector in the x-dir
Pachacha [2.7K]

Answer:

\vec{F}= -3.52\times 10^{-13}\hat{i}\ N

Explanation:

given,

charge = -5.0 μC

Electric force, F = 11 i^ N

force would a proton experience = ?

we know

\vec{F} = q \vec{E}

11\hat{i} = -5 \times 10^{-6} \vec{E}

\vec{E} =-2.2 \times 10^{6}\hat{i}

we know charge of proton is equal to 1.6 x 10⁻¹⁹ C

using formula

\vec{F} = q \vec{E}

\vec{F}= 1.6 \times 10^{-19}\times -2.2 \times 10^{6}\hat{i}

\vec{F}= -3.52\times 10^{-13}\hat{i}\ N

Force experienced by the photon in the same field is equal to \vec{F}= -3.52\times 10^{-13}\hat{i}\ N

3 0
2 years ago
Suppose a rock is dropped off a cliff with an initial speed of 0m/s. What is the rocks speed after 5 secounds, in m/s, if it enc
Tasya [4]

Answer:

The rock's speed after 5 seconds is 98 m/s.

Explanation:

A rock is dropped off a cliff.

It had an initial velocity of 0 m/s. And now it is moving downwards under the influence of gravitational force with the gravitational acceleration of 9.8 m/s².

Speed after 5 seconds = V

We know that acceleration = average speed/time

In our case,

g = ((0+V)/2)/5

9.8*5 = V/2

=> V = 2*9.8*5

V = 98 m/s

3 0
3 years ago
Read 2 more answers
Please help with these questions as well! I need urgent help! I will give brainliest! God bless!
Radda [10]

6.  Since we are not sure if the person in the question is actively lifting the crate, we have to determine the downwards force of the crate due to gravity and compare it to the normal force.  

F = ma

F = (15.3)(-9.8)

F = -150N

Since the downwards force of the crate is equivalent to the normal force, it means the person is applying no force in picking up the object.  So to pick up a 150N object from scratch, you would have to exert more force than the weight of the object, so the answer is 294N.


7.  Same idea as question 2.  

First determine the weight of the object:

F = ma

F = (30)(-9.8)

F = -294N

The crate in question is not moving, so the magnitudes of the forces in the upwards and downwards direction has to equal to 0.

-294 + 150N + x = 0

x = 144N  

So the person is exerting 144 N.


10.  First find the force of block B to the right due to its acceleration:

F = ma

F = (24)(0.5)

F = 12N

So block B is moving 12N to the right relative to block A due to block A's movement to the left.  However, block A is being applied a much greater force and is moving quicker to the left than block B is moving to the right of bock A.  The force that is causing block B to experience the lower relative force to the right is because of the friction.  To find the friction:

The sum of the forces in the leftward and rightward direction for block B must equal 12N.

75 - x = 12

x = 63N

So the force of friction of block A on block B is 63N to the left.


5 0
2 years ago
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