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Yuliya22 [10]
3 years ago
14

Three polarizing filters are stacked, with the polarizing axis ofthe second and third filters at angles of 22.2^\circ and 68.0^\

circ, respectively, to that of thefirst. If unpolarized light is incident on the stack, the light hasan intensity of 75.5 W/cm^2 after it passes through thestack.
a) If the incident intensity is kept constant, what is theintensity of the light after it has passed through the stack if thesecond polarizer is removed?
Physics
1 answer:
andreev551 [17]3 years ago
8 0

Answer:

I₂ = 25.4 W

Explanation:

Polarization problems can be solved with the malus law

     I = I₀ cos² θ

Let's apply this formula to find the intendant intensity (Gone)

Second and third polarizer, at an angle between them is

    θ₂ = 68.0-22.2 = 45.8º

    I = I₂ cos² θ₂

    I₂ = I / cos₂ θ₂

    I₂ = 75.5 / cos² 45.8

    I₂ = 155.3 W

We repeat for First and second polarizer

   I₂ = I₁ cos² θ₁

   I₁ = I₂ / cos² θ₁

   I₁ = 155.3 / cos² 22.2

   I₁ = 181.2 W

Now we analyze the first polarizer with the incident light is not polarized only half of the light for the first polarized

    I₁ = I₀ / 2

   I₀ = 2 I₁

   I₀ = 2 181.2

   I₀ = 362.4 W

Now we remove the second polarizer the intensity that reaches the third polarizer is

    I₁ = 181.2 W

The intensity at the exit is

    I₂ = I₁ cos² θ₂

    I₂ = 181.2 cos² 68.0

   I₂ = 25.4 W

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{\large{\bold{\rm{\underline{Given \; that}}}}}

★ A grey hound pursues a hare and takes 5 leaps for every 6 leaps of the hare, but 3 leaps of the hound are equal to 5 leaps of the hare.

{\large{\bold{\rm{\underline{To\; find}}}}}

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{\large{\bold{\rm{\underline{Solution}}}}}

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{\large{\bold{\rm{\underline{Full \; Solution}}}}}

\dashrightarrow  As it's given that a grey hound pursues a hare and takes 5 leaps for every 6 leaps of the hare, but 3 leaps of the hound are equal to 5 leaps of the hare.

 So firstly let us assume a metres as the distance covered by the hare in one leap.

Ok now let's talk about 5 leaps,.! As it's cleared that the hare cover the distance of 5a metres.

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Now the distance travelled by the hound in it's 5 leaps..!

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 Now let us compare the speed of the hound and the hare. Let us calculate them in the form of ratio..!

  • 25/3a = 6a

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