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nadezda [96]
3 years ago
6

Polarization: Unpolarized light passes through three ideal polarizing filters. The first filter is oriented with a horizontal tr

ansmission axis, the second one has its transmission axis at 30° from the horizontal, and the third filter has a vertical transmission axis. What percent of the light gets through this combination?
Engineering
1 answer:
GaryK [48]3 years ago
8 0

Answer:

the percentage of light that gets through this combination is 9.38

Explanation:

Given the data in the question;

Let us represent the incident unpolarized light with I_0.

So, the amount of light intensity passing through the first polarizer will be;

I_1 = I_0 / 2 ------ let this be equation 1

An the amount of light intensity passing through the second polarizer will be;

I_2 = I_1cos²θ

given that; the second one has its transmission axis at 30°

so, we substitute;

I_2 = I_1 × cos²( 30° )

I_2 = I_1 × 0.75

I_2 = 0.75I_1

from equation; I_1 = I_0 / 2

I_2 = 0.75( I_0 / 2 )

I_2 = 0.375I_0 .

Now, the amount of light intensity passing through the third polarizer will be;

I_3 = I_2cos² ( 90° - 30° )

I_3 = I_2 × cos²( 60° )

I_3 = I_2  × 0.25

we substitute

I_3 = 0.375I_0  × 0.25

I_3 = 0.09375I_0

∴ I_3/I_0 × 100 = 0.09375 × 100

⇒ 9.38%

Therefore, the percentage of light that gets through this combination is 9.38

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Answer:

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3 years ago
What is the lowest Temperature in degrees C?, In degrees K? in degrees F? in degrees R
Colt1911 [192]

Answer:

-273.16 °C

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0 °K

0 °R

Explanation:

The lowest temperature is the absolute zero.

Absolute zero is at 0 degrees Kelvin, or 0 degrees Rankine, because these are absolute scales that have their zero precisely at the absolute zero.

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Celsius scale has the same degree separation as the Kelvin scale, but the zero is separated by 273.16 degrees. Therefore the lowest temperature in the Celsius scale is -273.16 °C.

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3 0
3 years ago
A charge of 11.748 nC is uniformly distributed along the x-axis from −2 m to 2 m . What is the electric potential (relative to z
svp [43]

Answer:

electric potential  =  22.36 volt

Explanation:

given data

charge Q =  11.748 nC

distance d = 5 - 2 = 3 m

length = 2 + 2 = 4 m

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solution

electric potential is express as

electric potential  = \frac{Q}{4\pi \epsilon  _o L} ln(1+\frac{L}{d})   ..............1

electric potential  = \frac{KQ}{L} ln(1+\frac{L}{d})  

put here value

electric potential  = \frac{8.98755\times 10^9\times 11.748\times 10{-9}}{4} ln(1+\frac{4}{3})  

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6 0
3 years ago
We need to design a logic circuit for interchanging two logic signals. The system has three inputs I1I1, I2I2, and SS as well as
Salsk061 [2.6K]

Explanation:

Inputs and Outputs:

There are 3 inputs = I₁, I₂, and S

There are 2 outputs = O₁ and O₂

The given problem is solved in three major steps:

Step 1: Construct the Truth Table

Step 2: Obtain the logic equations using Karnaugh map

Step 3: Draw the logic circuit

Step 1: Construct the Truth Table

The given logic is

When S = 0 then O₁ = I₁ and O₂ = I₂

When S = 1 then O₁ = I₂ and O₂ = I₁

I₁     |     I₂     |    S    |    O₁    |    O₂

0     |     0     |    0    |    0    |     0

0     |     0     |    1     |    0    |     0

0     |     1      |    0    |    0    |     1

0     |     1      |    1     |    1     |     0

1      |     0     |    0    |    1     |     0

1      |     0     |    1     |    0    |     1

1      |     1      |    0    |    1     |     1

1      |     1      |    1     |    1     |     1

Step 2: Obtain the logic equations using Karnaugh map

Please refer to the attached diagram where Karnaugh map is set up.

The minimal SOP representation for output O₁

$ O_1 = I_1 \bar{S}  + I_2 S $

The minimal SOP representation for output O₂

$ O_2 = I_2 \bar{S}  + I_1 S $

Step 3: Draw the logic circuit

Please refer to the attached diagram where the circuit has been drawn.

7 0
3 years ago
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