If you increase the distance by double the work will _____________
1 answer:
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Answer: the wavelength of this transition is 1.2039 um
Explanation:
Given that;
the energy level between the transitioning energy gap Eg = 1.0 + 0.03 = 1.03 eV
we know that λ = 1.24 / Eg
so we substitute our Eg into the above equation
λ = 1.24 / 1.03
λ = 1.2039 um
therefore the wavelength of this transition is 1.2039 um
Answer:
the minimum wall thickness that will enhance the reflection of light is 146.9 nm
Explanation:
Given the data in the question;
At the first interface, a phase shift occurs as the incident light is in air that has less refractive index compare to the thin film of soap bubble.
At the second interface, no shift occurs,
condition for constructive interference;
t = ( m + 1/2) × λ/2n
where m = 0, 1, 2, 3 . . . . . .
now, the condition for the constructive interference;
t = mλ/2n
where t is the thickness of the soap bubble, λ is the wavelength of light and n is the refractive index of soap bubble.
so the minimum thickness of the film which will enhance reflection of light will be;
t = ( m + 1/2) × λ/2n
we substitute
t = ( 0 + 1/2) × 711 /2(1.21)
t = 0.5 × 711/2.42
t = 0.5 × 293.80165
t = 146.9 nm
Therefore, the minimum wall thickness that will enhance the reflection of light is 146.9 nm
Answer:
- The difference in length for steel is 2.46 x 10⁻⁴ m
- The difference in length for invar is 1.845 x 10⁻⁵ m
Explanation:
Given;
original length of steel, L₁ = 1.00 m
original length of invar, L₁ = 1.00 m
coefficients of volume expansion for steel, = 3.6 × 10⁻⁵ /°C
coefficients of volume expansion for invar, = 2.7 × 10⁻⁶ /°C
temperature rise in both meter stick, θ = 20.5°C
Difference in length, can be calculated as:
L₂ = L₁ (1 + αθ)
L₂ = L₁ + L₁αθ
L₂ - L₁ = L₁αθ
ΔL = L₁αθ
Where;
ΔL is difference in length
α is linear expansivity =
Difference in length, for steel at 20.5°C:
ΔL = L₁αθ
Given;
L₁ = 1.00 m
θ = 20.5°C
ΔL = 1 x 1.2 x 10⁻⁵ x 20.5 = 2.46 x 10⁻⁴ m
Difference in length, for invar at 20.5°C:
ΔL = L₁αθ
Given;
L₁ = 1.00 m
θ = 20.5°C
ΔL = 1 x 0.9 x 10⁻⁶ x 20.5 = 1.845 x 10⁻⁵ m