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KatRina [158]
2 years ago
6

what is the minimum (nonzero) thickness of the coating for which there is maximum transmission of the light into the rod?

Physics
1 answer:
timofeeve [1]2 years ago
4 0

The thickness is 155 nm at t1.

The thickness is 77.3 nm at t2.

The inquiry informs us that the laser light's wavelength is λ=510nm

The plastic rod's refractive index is n=1.30

The transparent coating's refractive index is nr=1.65

Minimum reflection would be required for maximal light transmission into the rod, and it is mathematically described as

2t1=510+10⁻⁹/1.65

t1=510+10⁻⁹/1.65*2

t1=155nm

where m is the interference order, which equals 1.

2t2= {m+1/2} λ/nr

The thickness is t replacing values

t1=155 nm

The highest reflection would occur for minimal light penetration through the rod, and this maximum reflection is mathematically described as

2t2= [m+1/2] λ/nr

t2=77.3 nm

The complete question is- Laser light of wavelength 510 nm is traveling in air and shines at normal incidence onto the flat end of a transparent plastic rod that has n = 1.30. The end of the rod has a thin coating of a transparent material that has refractive index 1.65. What is the minimum (nonzero) thickness of the coating (a) for which there is maximum transmission of the light into the rod; (b) for which transmission into the rod is minimized?

Learn more about reflection here-

brainly.com/question/15487308

#SPJ4

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Answer:

Δω = -5.4 rad/s

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Explanation:

<u>Given</u>:

           Initial angular velocity = ωi = 2.70 rad/s

           Final angular velocity = ωf = -2.70 rad/s (negative sign is  

           due to the movement in opposite direction)

           Change in time period = Δt = 1.50 s

<u>Required</u>:

           Change in angular velocity = Δω = ?

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<u>Solution</u>:

          <u>Angular velocity (Δω):</u>

               Δω = ωf - ωi

               Δω = -2.70 - 2.70

               Δω = -5.4 rad/s.

          <u> Average angular acceleration (αav):</u>

               αav = Δω/Δt

               αav = -5.4/1.50

              αav = -3.6 rad/s²

Since, the angular velocity is decreasing from 2.70 rad/s (in counter clockwise direction) to rest and then to -2.70 rad/s (in clockwise direction) so, the change in angular velocity is negative.

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A solid conducting sphere of radius 2 cm has a charge of 8microCoulomb. A conducting spherical shell of inner radius 4 cm andout
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Answer:

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Explanation:

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  • As the electric field must be zero inside the conducting spherical shell, this means that the charge enclosed by a spherical gaussian surface of a radius between 4 and 5 cm, must be zero too.
  • So, the +8 μC charge of  the solid conducting sphere of radius 2cm, must be compensated by an equal and opposite charge on the inner surface of the conducting shell of total charge -4 μC.
  • So, on the outer surface of the shell there must be a charge that be the difference between them:

        Q_{enc} = - 4e-6 C - (-8e-6 C) = + 4 e-6 C

  • Replacing in (1) A = 4*π*ε₀, and Qenc = +4 μC, we can find the value of E, as follows:

      E = \frac{1}{4*\pi*\epsilon_{0} } *\frac{Q_{enc} }{r^{2} } = \frac{9e9 N*m2/C2*4e-6C}{(0.07m)^{2} } = 7.35e6 N/C

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A mass of 0.250 kg is attached to a spring and undergoes simple harmonic oscillations with a period of 0.640 s. What is the forc
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The force constant of the spring is approximately 24.038 newtons per meter.

As we are talking about Simple Harmonic Motion. In this exercise we need to determine the Spring Constant (k), in newtons per meter, from the equation of the Period (T), in seconds, which is described below:

T = 2\pi\cdot \sqrt{\frac{m}{k} } (1)

Where m is the mass of the moving element, in kilograms.

If we know that T = 0.640\,s and m = 0.250\,kg, then the spring constant of the spring is:

0.640 = 2\pi\cdot \sqrt{\frac{0.250}{k} }

\sqrt{\frac{0.250}{k} } \approx 0.102

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The force constant of the spring is approximately 24.038 newtons per meter.

Please see this question related to Simple Harmonic Motion for further details: brainly.com/question/17315536

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