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2 years ago

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A beam of unpolarized light shines on a stack of five ideal polarizers, set up so that the angles between the polarization axes

of pairs of adjacent polarizers are all equal. The intensity of the transmitted beam is reduced from the intensity of the initial beam by a factor of phi=0.277 . Find the angle theta between the axes of each pair of adjacent polarizers.

1 answer:

ddd [48]2 years ago

7 0

Answer:

$21.75$

Explanation:

$n$ = number of polarizers through which light pass through = 5

$\theta$ = Angle between each pair of adjacent polarizers

$I_{o}$ = Intensity of unpolarized light

$I_{n}$ = Intensity of transmitted beam after passing all polarizers

It is given that

$\frac{I_{n}}{I_{o}}= 0.277$

we know that the intensity of light after passing through "n" polarizers is given as

$I_{n} = (0.5) I_{o} Cos^{2n-2}\theta$

$\frac{I_{n}}{I_{o}} = (0.5) Cos^{2n-2}\theta$

inserting the values

$0.277 = (0.5)Cos^{2(5)-2}\theta$

$0.554 = Cos^{8}\theta$

$Cos\theta = 0.9288$

$\theta = Cos^{-1}(0.9288)$

$\theta = 21.75$

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If the light strikes the first mirror at an angle θ1, what is the reflected angle θ2? express your answer in terms of θ1.

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Answer:

θ₂ = 90° - θ₁

Explanation:

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Since the light strikes the mirror at angle of θ₁, which is the angle between light ray and the mirror.

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2 years ago

Two speakers, A and B, produce identical sound waves. A listener is 3.2 m away from speaker A. The listener finds the lowest fre

earnstyle [38]

Answer:

0.83 m or 5.57 m

Explanation:

Destructive interference will occur when the distances from the speakers differ by 1/2 wavelength.

The length of 1 cycle of 72.4 Hz is ...

λ = v/f = (343 m/s)/(72.4 Hz) ≈ 4.738 m

So, the distance of the listener from speaker B is ...

3.2 m ± (4.738 m)/2 = {0.83 m, 5.57 m} . . . either of these distances

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The location could be at additional multiples of 4.738 m, but we think not. The sound intensity drops off with the square of the distance from the speaker, so identical sound waves from the speakers will sound quite different at different distances from the speakers. For best interference, the distances need to be as close to the same as possible. That will be at 3.2 m and 5.57 m.

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<em>Comment on the speed of sound</em>

We don't know what speed you are to use for the speed of sound. We have used 343 m/s. Some sources use 340 m/s, which will give a result different by 2 or 3 cm.

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When a test charge q0 = 2 nC is placed at the origin, it experiences a force of 8 times 10-4 N in the positive y direction. What

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Answer:

Electric field, $E=4\times 10^5\ N/C$

Explanation:

It is given that,

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wolverine [178]

Answer:

V

I and II

III and IV

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Taking always east as positive direction, and labelling

u the initial velocity

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(II)

u = 25 m/s

v = 0

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(III)

In this case,

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$\Delta t = 10 s$ is the time

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(IV)

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$\Delta t = 10 s$ is the time

So the magnitude of the impulse is

$|I| =F\Delta t = (2000N)(10)=20,000 Ns$

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So the ranking from largest to smallest is:

V

I and II

III and IV

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