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anyanavicka [17]
2 years ago
11

In your research lab, a very thin, flat piece of glass with refractive index 1.50 and uniform thickness covers the opening of a

chamber that holds a gas sample. The refractive indexes of the gases on either side of the glass are very close to unity. To determine the thickness of the glass, you shine coherent light of wavelength lambda 0 in vacuum at normal incidence onto the surface of the glass. When lambda 0 = 496 nm, constructive interference occurs for light that is reflected at the two surfaces of the glass. You find that the next shorter wavelength in vacuum for which there is constructive interference is 386 nm. Use these measurements to calculate the thickness of the glass. Express your answer with the appropriate units. What is the longest wavelength in vacuum for which there is constructive interference for the reflected light?
Physics
1 answer:
lara31 [8.8K]2 years ago
3 0

Answer:

The longest wavelength in vacuum for which there is constructive interference for the reflected light, λ   = 3472.

Explanation:

Refractive index of Glass (given) = 1.5

For the case of a constructive interference,

2nt = (m + 1/2) λ

For case 1,

2nt = (m + 1/2) 496 nm

For case 2,

2nt = (m +1+ 1/2) 386 nm

2nt = (m+3/2) * 386 nm

(m + 1/2) 496 nm = (m+3/2) * 386 nm

m = 3

Inserting the value of m in 1.

2nt = (m + 1/2) 496 nm

2*1.5t = (3 + 1/2) * 496 nm

t = ((3 + 1/2) * 496 nm)/ 3

t = 578.6 nm

The thickness of the glass, t = 578.6 nm

b)

It is generally known that for constructive interference,

2nt = (m + 1/2) λ

λ = 2nt / ((m + 1/2))

For Longest Wavelength, m = 0

λ = 2*1.5*578.6/ (1/2)

λ = 3472 nm

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Answer:

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Putting values, we get;

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3 years ago
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aleksklad [387]

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A. More than 20% of your daily recommended amount.

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3 years ago
Gold, which has a mass of 19.32 g for each cubic centimeter of volume, is the most ductile metal and can be pressed into a thin
Dennis_Churaev [7]

Answer:

area = 5733.33  cm²

length = 5.47 ×10^{7} cm

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to find out

area of the leaf and  length of the fiber

solution

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volume = mass / density

volume = 33.16 /  19.32

volume = 1.72 cm³

we know that volume = thickness × area

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area = volume / thickness

area = 1.72 / 3 ×10^{-4}

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frutty [35]

Answer:

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1 mm = 10^{-3} m

1 m = 10^{-3} km

So, 1 mm = 10^{-3}×10^{-3} km = 10^{-6} km.

As here the the distance is 4012 mm, then the distance in km will be

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So the distance is 4012 ×10^{-6} km.

5 0
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Answer:

m/s^2

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= kg m/s^2 ÷ kg

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