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Mrac [35]
4 years ago
9

An unpolarized light shines on an object that is placed below a crystal. When an observer looks at the object through the crysta

l there are two images present. Explain how that can happen using the information learned in optics and e-m radiation
Physics
1 answer:
slavikrds [6]4 years ago
8 0

Answer:

See explanation

Explanation:

Solution:-

- light reflected through calcite crystals difference occurs due to the polarity of the light passing through the crystal. Daylight is composed of light vibrating in all planes, whereas reflected light is often restricted to a single plane that is parallel to the surface from which the light is reflected.

- Light that is reflected from the flat surface of a dielectric (or insulating) material is often partially polarized, with the electric vectors of the reflected light vibrating in a plane that is parallel to the surface of the material.

- light waves that have the electric field vectors parallel to the surface are reflected to a greater degree than those with different orientations.

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An object is moving in a circle. If the radius of the object is doubled, and the period remains constant, the magnitude of the v
Anton [14]

Answer:

Twice

Explanation:

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What is the efficiency of a frictionless ramp that is 10 meters long and 1 meter high?
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3 years ago
Vector A has a magnitude of 6.0 m and points 30° north of east. Vector B has a magnitude of 4.0 m and points 30° west of south.
Nina [5.8K]

Answer:

The resultant vector \vec R = \vec A+\vec B is given by \vec R = 3.196\,\hat{i}-0.464\,\hat{j}\,\,\,[m].

Explanation:

Let \vec A = 6\cdot (\cos 30^{\circ}\,\hat{i}+\sin 30^{\circ}\,\hat{j}) and \vec B = 4\cdot (-\sin 30^{\circ}\,\hat{i}-\cos 30^{\circ}\,\hat{j}), both measured in meters. The resultant vector \vec R is calculated by sum of components. That is:

\vec R = \vec A+\vec B (Eq. 1)

\vec R = 6\cdot (\cos 30^{\circ}\,\hat{i}+\sin 30^{\circ}\,\hat{j})+4\cdot (-\sin 30^{\circ}\,\hat{i}-\cos 30^{\circ}\,\hat{j})

\vec R = (6\cdot \cos 30^{\circ}-4\cdot \sin 30^{\circ})\,\hat{i}+(6\cdot \sin 30^{\circ}-4\cdot \cos 30^{\circ})\,\hat{j}

\vec R = 3.196\,\hat{i}-0.464\,\hat{j}\,\,\,[m]

The resultant vector \vec R = \vec A+\vec B is given by \vec R = 3.196\,\hat{i}-0.464\,\hat{j}\,\,\,[m].

8 0
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MA_775_DIABLO [31]

Answer:

90 degrees

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vector A is pointing horizontally to the right, vector B (positive), points up vertically.

Now analyzing the subtraction of vectors A and B = A - B, we have that vector A continues to point to the right (positive) and vector B points down (negative). In this way the magnitudes of both operations sum and subtraction of vectors is equal.

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