Consider 3 polarizers. Polarizer 1 has a vertical transmission axis and polarizer 3 has a horizontal transmission axis. Taken to
gether, polarizers 1 and 3 are a pair of crossed polarizers. Polarizer 2, with a transmission axis of 300 to the vertical, is placed between polarizers 1 and 3. What is the intensity of light transmission through these polarizers given an initial intensity of light lo?
a. 0.375 lo
b. 0.25 l0
c. 0.18 lo
d. 0.09 lo
e. zero
a. 0.375 lo
b. 0.25 l0
c. 0.18 lo
d. 0.09 lo
e. zero
1 answer:
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Answer:
The new distance is d = 0.447 d₀
Explanation:
The electric out is given by Coulomb's Law
F = k q₁ q₂ / r²
This electric force is in balance with tension.
We reduce the charge of sphere B to 1/5 of its initial value (=q₂ = q₂ / 5) than new distance (d = n d₀)
dat
q₁ =
q₂ =
r = d₀
In order for the deviation to maintain the electric force it should not change, so we apply the Coulomb equation for the two points
F = k q₁ q₂ / d₀²
F = k q₁ (q₂ / 5) / (n d₀)²
.k q₁ q₂ / d₀² = q₁ q₂ / (5 n² d₀²)
5 n² = 1
n = √ 1/5
n = 0.447
The new distance is
d = 0.447 d₀