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Norma-Jean [14]
4 years ago
6

A beam of unpolarized light with intensity I0 falls first upon a polarizer with transmission axis θTA,1 then upon a second polar

izer with transmission axis θTA,2, where θTA,2−θTA,1=90degrees (in other words the two axes are perpendicular to one another). What is the intensity I2 of the light beam emerging from the second polarizer?
Physics
1 answer:
loris [4]4 years ago
5 0

Answer:

The intensity I₂ of the light beam emerging from the second polarizer is zero.

Explanation:

Given:

Intensity of first polarizer = Io/2

For the second polarizer, the intensity is equal:

I_{2} =\frac{I_{o} }{2} (cos\theta )^{2} =\frac{I_{o} }{2} (cos90)^{2} =0

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Three forces act on an object. Two of the forces have the magnitudes 56 N and 27 N, and make angles 53° and 156°, respectively,
Ludmilka [50]

Answer:

Explanation:

Given

Two forces F_1 and F_2 at an angle of \theta _1  \theta _2

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\theta _2=156^{\circ}

As resultant force is zero therefore horizontal component as well as vertical component of force is zero

\sum F_x=F_1\cos \theta _1+F_2\cos \theta _2+F_3\cos \theta _3

\sum F_x=56\cos 53+27\cos 156+F_3\cos \theta

F_3\cos \theta_3=-9.035\ N----1

\sum F_y=F_1\sin \theta _1+F_2\sin \theta _2+F_3\sin \theta _3

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squaring and adding 1 and 2

F_3^2(\cos ^2\theta _3+\sin^2\theta _3)=86.583+3102.93

F_3=56.47\ N

Divide 1 and 2 to get \theta _3

\frac{\sin \theta_3 }{\cos \theta_3 }=\frac{-55.704}{9.305}

\tan \theta_3=-6.16

\theta _3=180-80.78

\theta _3=99.22^{\circ}

 

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