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Readme [11.4K]
3 years ago
11

"A thin film with an index of refraction of 1.50 is placed in one of the beams of a Michelson interferometer. If this causes a s

hift of 8 bright fringes in the pattern produced by light of wavelength 540 nm, what is the thickness of the film?"
Physics
1 answer:
Tamiku [17]3 years ago
6 0

Answer:

The film thickness is 4.32 * 10^-6 m

Explanation:

Here in this question, we are interested in calculating the thickness of the film.

Mathematically;

The number of fringes shifted when we insert a film of refractive index n and thickness L in the Michelson Interferometer is given as;

ΔN = (2L/λ) (n-1)

where λ is the wavelength of the light used

Let’s make L the subject of the formula

(λ * ΔN)/2(n-1) = L

From the question ΔN = 8 , λ = 540 nm, n = 1.5

Plugging these values, we have

L = ((540 * 10^-9 * 8)/2(1.5-1) = (4320 * 10^-9)/1 = 4.32 * 10^-6 m

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Answer:

A) Object A is 3.25 times hotter.

B) Object A radiates 111.6 times more energy per unit of area.

Explanation:

Wiens's law states that there is an inverse relationship between the wavelength in which there is a peak in the emission of a black body and its temperature, mathematically,

\lambda_{peak}= \dfrac{0.0028976}{T},

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For object A:

T_{A}=\dfrac{0.0028976}{200*10^{-9}}

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T_{B}=\dfrac{0.0028976}{650*10^{-9}}

T_{B}=4458K

From this, we get that

T_{A}/T_{B}=3.25,

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Stefan's Law states that a black body emits thermal radiation with power proportional to the fourth power of its temperature.

This is

E=\sigma T^{4},

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From this, power can be easily compute:

E_{A}=(5.67*10^{-8}*(14488)^{4})=2.5*10^{9}W\\E_{B}=(5.67*10^{-8}*(4458)^{4})=22.4*10^{6}}W,

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E_{A}/E_{B}=111.6,

which means that object A radiates 111.6 time more energy per unit of area.

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Then we can replace these two values in the equation to get:

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