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Harrizon [31]
3 years ago
12

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.
Physics
1 answer:
ddd [48]3 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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Answer:

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     The  initial magnetic field is  B_i  =  0.11 \ T

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    The time taken is  t  = 12 \ ms  =  12*10^{-3} \  s

   

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substituting values

         r =  \frac{0.022}{2}

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Generally the induced emf  is mathematically represented as

      \epsilon  =  - N *  \frac{d\phi}{dt}

Where  d\phi is the change in magnetic flux of the wire which is mathematically represented as

      d \phi  =  dB*  A *  cos \theta

=>  d \phi  =  (B_f  -  B_i )*  A *  cos \theta

Here  \theta  =  0

since the axis of the coil is parallel to the field

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      A  =  3.80*10^{-4} \  m^2

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       \epsilon  = 4.53 \  V

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Answer:

please find the attachment to this question.

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