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Debora [2.8K]
3 years ago
11

Imagine that you have a very thin uniform oil film on the surface of water. The thickness of oil is much smaller than the wavele

ngth of blue light. White light is shining on the film. What color would the film be if you looked at it from above? Explain.
Physics
1 answer:
Leona [35]3 years ago
5 0

Answer:

Green light

Explanation:

The thickness of oil is smaller than wavelength of blue light. The wavelength 450 nm to 495 nm. Let the thickness of oil is 400 nm.

By the constructive interference formula,

2nt = m\lamda\lamda\\λ

2 X 1.28 X 400 X 10^(-9) = 2λ

λ = 512 nm

This is the wavelength of green light.

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a * 1.2 = -16.0 + 9.50
a * 1.2 = - 6.5 
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d = 16.0 * 1.2 - ( 5.4167 * 1.2² / 2 ) =
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When a car moves up a hill with constant
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The origin of an x axis is placed at the center of a nonconducting solid sphere of radius R that carries a charge +qsphere distr
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Answer:

q=49Q/64

and

x =16R/15

Explanation:

See  attached figure.

E_{Q}= E due to sphere

E_{q}= E due to particule

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according to the law of gauss and superposition Law:

E_{Q}=E_{1}+E_{2}=E_{2} ; electric field due to the small sphere with r1=R/4

E_{Q}=kq_{2}/(r_{1}^{2})=

q_{2}=density*4/3*pi*r_{1}^{3}=Q/(4/3*pi*R^{3})*4/3*pi*r_{1}^{3}=Q*r_{1}^{3}/R^{3}

then: E_{Q}=kq_{2}/(r_{1}^{2})=k*Q*r_{1}^{3}/(R^{3}*r_{1}^{2}) = kQ/(4*R^{2})  (2)

on the other hand, for the particule:

E_{q}=kq/(r_{p}^{2})

r_{p}=2R-R/4=7R/4   ⇒    E_{q}=16kq/(49R^{2})   (3)

We replace (2) y (3) in (1):

E_{total}=E_{Q}-E_{q}=0=kQ/(4*R^{2}) - 49kq/(16R^{2})

q=49Q/64

--------------------

if R<x<2R   AND E_{total}=E_{Q}-E_{q}=0

E_{total}=E_{Q}-E_{q}=0=kQ/(x^{2}) - kq/(2R-x^{2})

remember that  q=49Q/64

then:

Q(2R-x^{2})=49/64*x^{2}

solving:

x_{1} =16R/15

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7 0
3 years ago
Any ss2 here (11th Grade)
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Answer:

<h2>a) 50°</h2><h2>b) 40°</h2>

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Check the complete diagram n the attachment below

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Given angle of glance on M₂ to be 40°, r₂ = 90-40 = 50°

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The angle of incidence on M₁ = 90° - 40° = 50°

b) The angle of incidence to the surface of M₁(∠PO₁A)will be the angle of glance on M₁ which is equivalent to 40°

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