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ad-work [718]
3 years ago
14

Unpolarized light of intensity I=10 falls on two successive polarizer wheels with the angle between the polarizer wheels e-60°.

What is the resultant intensity of the polarized light after leaving the 2nd polarizer? A) (0)/2 B) (10)/4 C) (0)/8 D) (o)/16
Physics
1 answer:
yan [13]3 years ago
5 0

Answer:

I=\frac{10}{4}

Explanation:

A polarizer changes the orientation of the oscillations of a light wave.

I₀ = Intensity of unpolarized light = 10

θ = Angle given to the polarizer = 60°

Intensity of light

I = I₀cos²θ

⇒I = 10cos²60

\\\Rightarrow I=10\times \frac{1}{2}\times \frac{1}{2}\\\Rightarrow I=\frac{10}{4}

So, the after passing through the second polarizer is \mathbf{\frac{10}{4}}

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shows two parallel nonconducting rings with their central axes along a common line. Ring 1 has uniform charge q1 and radius R; r
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\dfrac{1}{4\pi\epsilon _o}\dfrac{q_1R}{(R^2+R^2)^{3/2}}-\dfrac{1}{4\pi\epsilon _o}\dfrac{q_2(d-R)}{(R^2+(d-R)^2)^{3/2}}=0

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8 0
3 years ago
Read 2 more answers
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