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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Answer:
Option C - the angle of refraction is greater than the angle of incidence
Explanation:
Snell's law of refraction states that;
n1 sinθ1 = n2 sinθ2
Where;
n1 is refractive index of incidence medium
θ1 is angle of incidence
n2 is Refractive index of refraction medium
θ2 is angle of refraction
For This question, n1 = 1.5 and n2 = 1.33
Thus;
1.5 sinθ1 = 1.33 sinθ2
Rearranging, we have;
sinθ1/sinθ2 = 1.33/1.5
We know from trigonometry, that sin 0 = 0 and sin 90 = 1. So, as θ approaches 0°, the value of sinθ decreases while as it approaches 90°,the value of sinθ increases.
Thus, by inspection, we can say that the value of the denominator is higher than the numerator.
Thus, θ2 is greater than θ1
So, the angle of refraction is greater than the angle of incidence
Answer:
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