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Olegator [25]
3 years ago
10

Unpolarized light with intensity S is incident on a series of polarizing sheets. The first sheet has its transmission axis orien

ted at 0°. A second polarizer has its transmission axis oriented at 45° and a third polarizer oriented with its axis at 90°. Determine the fraction of light intensity exiting the third sheet with and without the second sheet present.
Physics
1 answer:
jeka943 years ago
3 0

Answer:

Explanation:

Given

Initial Intensity of light is S

when an un-polarized light is Passed through a Polarizer then its intensity reduced to half.

When it is passed through a second Polarizer with its transmission axis \theat =45^{\circ}

S_1=S_0\cos ^2\theta

here S_0=\frac{S}{2}

S_1=\frac{S}{2}\times \frac{1}{(\sqrt{2})^2}

S_1=\frac{S}{4}

When it is passed through third Polarizer with its axis 90^{\circ} to first but \theta =45^{\circ} to second thus S_2

S_2=S_0\cos ^2\theta

S_2=\frac{S}{4}\times \frac{1}{2}

S_2=\frac{S}{8}

When middle sheet is absent then Final Intensity will be zero                    

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artcher [175]
<h2>Answer:</h2>

0.126m

<h2>Explanation:</h2>

According to Hooke's law, the force (F) acting on a spring to cause an extension or compression (e) is given by;

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Where;

k = the spring's constant.

From the question, the force acting on the spring is the weight(W) of the mass. i.e

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<em>But;</em>

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<em>From equation (ii), it implies that;</em>

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<em>Now substitute F = m x g into equation(i) as follows;</em>

F = k x e

m x g = k x e      ------------------(iii)

<em>From the question;</em>

m = m1 = 3.5kg

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3.5 x 10 = 278 x e

35 = 278e

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e = 35/278

e = 0.126m

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At its maximum height, some distance y above the point where the body is launched, the body has zero velocity, so

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Solve for y :

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At the highest point, we have

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t = (20 m/s) / (9.8 m/s²) ≈ 2.04 s

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Speed is the magnitude of velocity, so the speed in question is 39.7 m/s.

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