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Elden [556K]
3 years ago
14

A light beam strikes a piece of glass at a 65.00 ∘ incident angle. The beam contains two wavelengths, 450.0 nm and 700.0 nm, for

which the index of refraction of the glass is 1.4831 and 1.4754, respectively.
What is the angle between the two refracted beams?
Physics
2 answers:
Leokris [45]3 years ago
6 0

Answer:

Explanation:

Lower the refractive index ,  higher the wave length

Refractive index becoming high reduces the velocity and hence wavelength as frequency remains unchanged.

So refractive index  1.4831 will correspond to wavelength of 450 nm.

For  refractive index is 1.4831 , angle of incidence i , angle of refraction r .

Sin i / Sinr = 1.4831

sin65 / sinr = 1.4831

sir r = sin65 / 1.4831

= .9063 / 1.4831

= .6111

r = 37.67 degree

For  refractive index is 1.4754 , angle of incidence i

Sin i / Sinr = 1.4754

sin65 / sinr = 1.4754

sir r = sin65 / 1.4754

= .9063 / 1.4754

= .61427

r = 37.9 degree

angle between two refracted ray

37.9 -37.67

= 0. 23 degree

bagirrra123 [75]3 years ago
3 0

Answer:

0.2°

Explanation:

Angle of incidence on glass, i = 65°

first wavelength, λ1 = 450 nm

second wavelength, λ1 = 700 nm

first refractive index, μ1 = 1.4831

second refractive index, μ1 = 1.4754

Let the angle of refraction for the first wavelength is r1 and for the second wavelength is r2.

Use Snell's law

\mu_{1}=\frac{Sin i}{Sin r_{1}}

Sin r_{1}=\frac{Sin 65}{1.4831}

Sin r1 = 0.61109

r1 = 37.7°

\mu_{2}=\frac{Sin i}{Sin r_{2}}

Sin r_{2}=\frac{Sin 65}{1.4754}

Sin r2 = 0.6143

r2 = 37.9°

Angle between the two refracted beams = r2 - r1 = 37.9° - 37.7° = 0.2°

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Answer:

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As the potential and kinetic energies are conserved

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Ann [662]

Answer:

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4 0
2 years ago
How do I go about this?
Anna71 [15]

Hi there!

(a)

Recall that:
W = F \cdot d = Fdcos\theta

W = Work (J)
F = Force (N)
d = Displacement (m)

Since this is a dot product, we only use the component of force that is IN the direction of the displacement. We can use the horizontal component of the given force to solve for the work.

W =248(56)cos(30) = 12027.36 J

To the nearest multiple of ten:
W_A = \boxed{12030 J}

(b)
The object is not being displaced vertically. Since the displacement (horizontal) is perpendicular to the force of gravity (vertical), cos(90°) = 0, and there is NO work done by gravity.

Thus:
\boxed{W_g = 0 J}

(c)
Similarly, the normal force is perpendicular to the displacement, so:
\boxed{W_N = 0 J}

(d)

Recall that the force of kinetic friction is given by:
F_{f} =\mu_k mg

Since the force of friction resists the applied force (assigned the positive direction), the work due to friction is NEGATIVE because energy is being LOST. Thus:
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In multiples of ten:
\boxed{W_f = -3070 J}

(e)
Simply add up the above values of work to find the net work.

W_{net} = W_A + W_f \\\\W_{net} = 12027.36 + (-3073.28) = 8954.08 J

Nearest multiple of ten:
\boxed{W_{net} = 8950 J}}

(f)
Similarly, we can use a summation of forces in the HORIZONTAL direction. (cosine of the applied force)
F_{net} = F_{Ax} - F_f

W = F_{net} \cdot d = (F_{Ax} - F_f)

W = (F_Acos(30) - \mu_k mg)d\\W = (248cos(30) - 0.1(56)(9.8)) * 56 \\\\W = 8954.08 J

Nearest multiple of ten:
\boxed{W_{net} = 8950 J}

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Answer:

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Explanation:

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