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amid [387]
3 years ago
7

A scientist notices that an oil slick floating on water when viewed from above has many different colors reflecting off the surf

ace, making it look rainbow-like (an effect known as iridescence). She aims a spectrometer at a particular spot and measures the wavelength to be 750 nmnm (in air). The index of refraction of water is 1.33.
a. The index of refraction of the oil is 1.20. What is the minimum thickness ttt of the oil slick at that spot?
b. Suppose the oil had an index of refraction of 1.50. What would the minimum thickness be now?
c. Now assume that the oil had a thickness of 200 nm and an index of refraction of 1.5. A diver swimming underneath the oil slick is looking at the same spot as the scientist with the spectromenter. What is the longest wavelength of the light in water that is transmitted most easily to the diver?
Physics
1 answer:
Elan Coil [88]3 years ago
5 0

Answer:

(a) 313nm

(b) 250nm

(c) 301nm

Explanation:

This problem involves the concept of thin-film interference. The scientist observes constructive interference. In this case m = 1.

(a) To find the thickness of the oil at that spot we need to first calculate the wavelength in the oil.

       λ = λo/n = 750/ 1.20 = 625nm

λ = wavelength in the material of interest into which light is entering (the oil)

λo = wavelength in the material from which light is coming in or entering.

n = refractive index of the material into which the light is entering.

Light travels from air into the oil. So the refractive index used above is that of the oil and not air.

The formula relating the thickness t to the wavelength is 2t = mλ

m = 1 so,

    2t = λ = 625nm

     t = 625/2 = 313nm

(b) Supposing the refractive index of the water is now n = 1.50

The wavelength in the oil λ = 750/1.50 = 500nm

 The thickness t = λ/2 = 500/2 = 250nm

(c) If the oil has a thickness of 200nm then the wavelength λ = 2t = 2 ×200nm = 400nm

The wavelength in the oil is 400nm. In order to find the wavelength in water we set this wavelength to λo = 400nm. So the wavelength in water seen by the diver is given by the formula

       λ = λo/n

n for water is 1.33

      λ = 400/1.33 = 300.8nm ≈ 301nm.

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Three beads are placed along a thin rod. The first bead, of mass m1 = 23 g, is placed a distance d1 = 1.1 cm from the left end o
Mila [183]

Answer:

a) x=\frac{m_{1}d_{1}+m_{2}(d_{1}+d_{2})+m_{3}(d_{1}+d_{2}+d_{3}  ) }{m_{1}+m_{2}+m_{3} }

b) x = 4.47 cm

c) x=\frac{m_{1}d_{2}+m_{2}(0)+m_{3}d_{3} }{m_{1}+m_{2}+m_{3} }

d) x = 1.48 cm

Explanation:

a) The center of mass is equal to:

x=\frac{m_{1}x_{1}+m_{2}x_{2}+m_{3}x_{3} }{m_{1}+m_{2} +m_{3}}

Where m is the mass of beads and x is the distances, if x₁ = d₁, x₂ = d₂ and x₃ = d₃

x=\frac{m_{1}d_{1}+m_{2}(d_{1}+d_{2})+m_{3}(d_{1}+d_{2}+d_{3}  ) }{m_{1}+m_{2}+m_{3} }

b) If

m₁ = 23g

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d₁ = 1.1 cm

d₂ = 1.9 cm

d₃ = 3.2 cm

x=\frac{23*1.1+15*(1.1+1.9)+58(1.1+1.9+3.2) }{23+15+58 } =4.47cm

c) The center of the mass of the beads realtive to the center of bead is:

x=\frac{m_{1}d_{2}+m_{2}(0)+m_{3}d_{3} }{m_{1}+m_{2}+m_{3} }

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consider the motion in x-direction

v_{ox} = initial velocity in x-direction = ?

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a_{x} = acceleration along x-direction = 0 m/s²

t = time of travel = 4.60 sec

Using the equation

X = v_{ox} t + (0.5) a_{x} t²

100 =  v_{ox} (4.60)

v_{ox} = 21.7 m/s


consider the motion along y-direction

v_{oy} = initial velocity in y-direction = ?

Y = vertical displacement  = 0 m

a_{y} = acceleration along x-direction = - 9.8 m/s²

t = time of travel = 4.60 sec

Using the equation

Y = v_{oy} t + (0.5) a_{y} t²

0 = v_{oy} (4.60) + (0.5) (- 9.8) (4.60)²

v_{oy} = 22.54 m/s

initial velocity is given as

v_{o} = sqrt((v_{ox})² + (v_{oy})²)

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A large container contains a large amount of water. A hole is drilled on the wall of the container, at a vertical distance h = 0
barxatty [35]

Answer:

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This problem can be solved if we use the Bernoulli equation: In the attached image we can see the conditions of the water inside the container.

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The velocity in the point 1 is zero because we have this conditional statement "The water surface drops very slowly and its speed is approximately zero"

h2 is located at point 2 and it will be zero.

(P_{1} +\frac{v_{1}^{2} }{2g} +h_{1} )=(P_{2} +\frac{v_{2}^{2} }{2g} +h_{2} )\\P_{1} =P_{2} \\v_{1}=0\\h_{2} =0\\v_{2}=\sqrt{0.54*9.81*2}\\v_{2}=3.25[m/s]

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CaHeK987 [17]

B.riding on a ferris wheel

E.a satellite orbiting the Earth

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A person riding on a ferris wheel and a satellite orbiting the earth both depicts circular motion.

Circular motion is the motion of a body about a fixed axis or center.

  • The motion path of the body forms a circle around the center.
  • In circular motion, the travel path is along a circle.
  • An orbiting satellite traces out its orbit round the earth.
  • At certain times, it passes through that point over and over again.
  • This is similar to that of a ferris wheel

Other options gives other types of motion.

learn more:

circular motion brainly.com/question/2562955

#learnwithBrainly

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