21.13 The index of refraction of corundum (Al2O3) is anisotropic. Suppose that visible light is passing from one grain to anothe
1 answer:
Answer:
3.87 x 10⁻⁵
Explanation:
Given parameters
normal incidence
refraction index of grain1, n₁ = 1.757
refraction index of grain2, n₂ = 1.779
to calculate reflectivity index between the two grains of different orientation an at normal incidence, we use the relation
since the incidence is normal where R, is the index of relativity
R = 3.87 x 10⁻⁵
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Answer: Inherent width in the emission line: 9.20 × 10⁻¹⁵ m or 9.20 fm
length of the photon emitted: 6.0 m
Explanation:
The emitted wavelength is 589 nm and the transition time is ∆t = 20 ns.
Recall the Heisenberg's uncertainty principle:-
∆t∆E ≈ h ( Planck's Constant)
The transition time ∆t corresponds to the energy that is ∆E
.
The corresponding uncertainty in the emitted frequency ∆v is:
∆v= ∆E/h = (5.273*10^-27 J)/(6.626*10^ J.s)= 7.958 × 10^6 s^-1
To find the corresponding spread in wavelength and hence the line width ∆λ, we can differentiate
λ = c/v
dλ/dv = -c/v² = -λ²/c
Therefore,
∆λ = (λ²/c)*(∆v) = {(589*10⁻⁹ m)²/(3.0*10⁸ m/s)} * (7.958*10⁶ s⁻¹)
= 9.20 × 10⁻¹⁵ m or 9.20 fm
The length of the photon (<em>l)</em> is
l = (light velocity) × (emission duration)
= (3.0 × 10⁸ m/s)(20 × 10⁻⁹ s) = 6.0 m
Answer:
Temperature at center of apples = 11.2⁰C
Temperature at surface of apples = 2.7⁰C
Amount of Heat transferred = 17.2kJ
Explanation:
The properties of apple are given as:
k = 0.418 W/m.°C
ρ = 840 kg/m³
Cр = 3.81 kJ/kg.°C
α = 1.3*10 ⁻⁷ m²/s
h = 8 W/m².°C
d = 0.09m
r = 0.045m
t = 1 hour = 3600s
<h2>Solution</h2>Biot number is given as:
The constants λ₁ and A₁ corresponding to Biot number (from the table) are:
λ₁ = 1.476
A₁ = 1.239
Fourier Number is:
As Fourier Number > 0.2 , one term approximates solutions are applicable
The temperature at the center of apples, The temperature at surface of apples and Amount of heat transfer is found in the ATTACHMENT.
Answer:
The solution for the given problem is done below.
Explanation:
M1 = 2.0
= 0.3636
= 0.5289
= 0.7934
Isentropic Flow Chart: M1 = 2.0 , = 1.8
T1 = (1.8)(288K) = 653.4 K.
In order to choke the flow at the exit (M2=1), the above T0* must be stagnation temperature at the exit.
At the inlet,
T02= = (1.8)(288K) = 518.4 K.
Q= Cp(T02-T01) = = 135.7*
J/Kg.