A21-w horizontal beam of light of wavelength 430 nm, travelling at speed c, passes through a rectangular opening of width 0.0048 m and height 0.011 m. the light then strikes a screen at a distance 0.36 m behind the opening.

a) calculate the energy, in joules, of a single photon in the beam.
b) find the number of photons emitted per second by the light source.
c) if the beam of light emitted by the source has a constant circular cross section whose radius is twice the height of the opening the beam is approaching, find the flow density of photons as the number of photons passing through a square meter of cross-sectional area per second.
d) calculate the number of photons that pass through the rectangular opening per second.
e) the quantity no that you found in part (d) gives the rate at which photons enter the region between the opening and the screen. assuming no reflection from the screen, find the number of photons in that region at any time.
f) how would the number of photons in the region between the opening and the screen change, if the photons traveled more slowly? assume no change in any other quantity, including the speed of light.

a)  E = 4.626 10⁻¹⁹ J , b)  # = 8.86 10¹⁹ photons , c)  σ = 5.83 10²² photon / m², d)    # photon = 3.078 10¹⁹ photons , e)  # photon = 3.078 10¹⁹ photon / s , f) The number of photons increases

Explanation:

a) For this part let's use the Planck equation

     E = h f

The speed of light is

    c = λ f

We replace

    E = h c /λ

Let's reduce to SI units

   λ = 430 nm (1m / 10⁹ nm) = 430 10⁻⁹ nm

    E = 6.63 10⁻³⁴4 3 10⁸ / 4.30 10⁻⁷

    E = 4.626 10⁻¹⁹ J

b) the defined power

   P = W / t = E / t

Let's look for the energy of the beam in 1 s

    E = P t

    E = 41 1 = 41 J

Now let's use a rule of proportions (rule of three) if 1 photon has an energy E₀ how many photons has an energy E

    #photons = E / E₀ 1 photon

   # photon = 41 / 4,626 10⁻¹⁹

   # = 8.86 10¹⁹ photons

c) The radius of the beam is

    r = 2 h

    r = 2 0.011

    r = 0.022 m

Density is

    σ = # Photons / A

The area of a circle is

   A = π R²

   σ. = # Photon /π r²

   σ = 8.86 1019 /π 0.022²

   σ = 5.83 10²² photon / m²

d) let's look for the opening area

   A₁ = l a

   A₁ = 0.0048 0.011

   A₁ = 5.28 10⁻⁴ m²

Let's use the density decision

    σ = # Photons / A

    # photon =σ . A1

   # photon = 5.83 10²² 5.28 10⁻⁵

  # photon = 3.078 10¹⁹ photons

e) let's look for the time it takes for photons to reach the screen.

The number of photons is the amount of photons per unit of bone time is equivalent to a speed

      v = # photon = d / t

      t = d / # foton

      t = 0.36 / 3.078 10¹⁹

      t = 1.17 10⁻²⁰ s

As there are no photons reflected on the screen, everything is absorbed, so the number of photons that enter is equal to the number of photons that are absorbed, therefore, the amount of photons does not change

         # photon = 3.078 10¹⁹ photon / s

f) If the speed of the photons decreases in the space between the aperture and the screen, an imbalance is created

More photons enter than are absorbed, so the number of photons increases the region. In order to calculate this number, the index of refraction of the medium or the speed of supply of this medium must be granted.

The number of photons increases

a charge 3q is at the origin, and a charge −2q is on the positive x axis at x=a. part a where on the x-axis would you place a third charge so it would experience no net electric force? enter an expression for the exact answer, which will involve a square-root sign. (do not enter an approximate decimal answer.)

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