1answer.
Ask question
Login Signup
Ask question
All categories
  • English
  • Mathematics
  • Social Studies
  • Business
  • History
  • Health
  • Geography
  • Biology
  • Physics
  • Chemistry
  • Computers and Technology
  • Arts
  • World Languages
  • Spanish
  • French
  • German
  • Advanced Placement (AP)
  • SAT
  • Medicine
  • Law
  • Engineering
Stolb23 [73]
3 years ago
15

In the macroscopic world, you know that you can hear but cannot see around corners. Under what conditions does light bend around

corners (i.e. diffract) ? Explain why sound diffracts easily around a classroom door. 5. Suppose you added to the single slit an identical slit a distance d=0.25mm away from the first. Draw the resulting interference pattern you might expect on the same screen. What happens when we increase the distance between slits ? What happens in the limit that d becomes arbitrarily large?
Physics
1 answer:
omeli [17]3 years ago
6 0

Answer:

a much larger slit, the phenomenon of Sound diffraction that slits for light.

this is a series of equally spaced lines giving a diffraction envelope

Explanation:

The diffraction phenomenon is described by the expression

    d sin θ = m λ

Where d is the distance of the slit, m the order of diffraction that is an integer and λ the wavelength.

 

For train the diffraction phenomenon, the d / Lam ratio is decisive if this relation of the gap separation in much greater than the wavelength does not reduce the diffraction phenomenon but the phenomena of geometric optics.

The wavelength range for visible light is 4 10⁻⁷ m to 7 10⁻⁷ m. The wavelength range for sound is 17 m to 1.7 10⁻² m. Therefore, with a much larger slit, the phenomenon of Sound diffraction that slits for light.

When we add a second slit we have the diffraction of each one separated by the distance between them, when the integrals are made we arrive at the result of the interference phenomenon, a this is a series of equally spaced lines giving a diffraction envelope

When I separate the distance between the two slits a lot, the time comes when we see two individual diffraction patterns

You might be interested in
What happens to the current in a circuit when the resistance is increased
VikaD [51]

then the current will decrease.

hope this helps :)

5 0
2 years ago
How would the carrying capacity for northern cardinals in a mixed-oak forest be affected if the number of nesting sites decrease
astraxan [27]

Answer:

They would decline

Explanation:

They would either migrate, or die.

3 0
3 years ago
Read 2 more answers
calculate the momentum of 5.07 kg egg that is dropped from a roof and falls 2 seconds before hitting the ground.
inysia [295]
Anything that's dropped on or near the Earth accelerates at the rate of
9.8 m/s² on its way down.  The mass of the object makes no difference. 
After falling for 2 seconds, it's falling at the rate of 19.6 m/s.

Momentum = (mass) x (speed)

The egg's momentum is   (5.07 kg) x (19.6 m/s) = 99.4 kg-m/s .
8 0
3 years ago
A pendulum is used in a large clock. The pendulum has a mass of 2 kg. If the pendulum is moving at a speed of 2.9 m/s when it re
Rufina [12.5K]
This is a classic example of conservation of energy. Assuming that there are no losses due to friction with air we'll proceed by saying that the total energy mus be conserved.
E_m=E_k+E_p
Now having information on the speed at the lowest point we can say that the energy of the system at this point is purely kinetic:
E_m=Ek=\frac{1}{2}mv^2
Where m is the mass of the pendulum. Because of conservation of energy, the total energy at maximum height won't change, but at this point the energy will be purely potential energy instead.
E_m=E_p
This is the part where we exploit the Energy's conservation, I'm really insisting on this fact right here but it's very very important, The totam energy Em was
E_M=\frac{1}{2}mv^2
It hasn't changed! So inserting this into the equation relating the total energy at the highest point we'll have:
E_p=mgh=E_m=\frac{1}{2}mv^2
Solving for h gives us:
h=\frac{v^2}{2g}.
It doesn't depend on mass!

4 0
3 years ago
Read 2 more answers
A 10-kg dog is running with a speed of 5.0 m/s. what is the minimum work required to stop the dog in 2.40 s?
ankoles [38]
Given required solution

M=10kg W=? W=Fd
v=5.0m/s F=mg
t=2.40s =10*10=100N
S=VT
=5m/s*2.4s
=12m
so W=12*100
W=1200J
4 0
2 years ago
Other questions:
  • What is the working principle of lever​
    11·1 answer
  • What units do physicists use to measure heat energy and electrical energy?
    13·1 answer
  • Calculate the period (T) of uniform circular motion if the velocity is 40.0 m/s and centripetal acceleration is 20.0 m/s2.
    10·1 answer
  • A uniform electric field of magnitude 6.8 × 10 5 N/C points in the positive x direction. (a) Find the electric potential differe
    6·1 answer
  • NEED HELP!!<br><br> To complete its outermost shell, oxygen will most likely ____
    7·2 answers
  • A boy flying a kite is standing 30 ft from a point directly under the kite. if the string to the kite is 50 ft long, what is the
    11·1 answer
  • Two imaginary spherical surfaces of radius R and 2R respectively surround a positive point charge Q located at the center of the
    7·1 answer
  • changes in which the form or shape of a substance changes but the substance keeps its identity are called
    9·1 answer
  • What type of energy does this wave carry?​
    14·1 answer
  • I need to catch up on late work
    13·1 answer
Add answer
Login
Not registered? Fast signup
Signup
Login Signup
Ask question!