Please help on this one?

Unpolarized light with intensity 370 W/m2 passes first through a polarizing filter with its axis vertical, then through a second

vampirchik [111]

To solve this problem it is necessary to apply the concepts given by Malus regarding the Intensity of light.

From the law of Malus intensity can be defined as

$I' = \frac{I_0}{2} cos^2 \theta$

Where

$\theta =$ Angle From vertical of the axis of the polarizing filter

$I_0 =$ Intensity of the unpolarized light

The expression for the intensity of the light after passing through the first filter is given by

$I = \frac{I_0}{2}$

Replacing we have that

$I = \frac{370}{2}$

$I = 185W/m^2$

Re-arrange the equation,

$I'= \frac{I_0}{2}cos^2\theta$

Re-arrange to find \theta

$cos^2\theta = \frac{2I'}{I_0}$

$cos^2\theta = \frac{2*138}{370}$

$\theta = cos^{-1}(\sqrt{\frac{2*138}{370}})$

$\theta = 0.5282rad$

$\theta = 30.27\°$

The value of the angle from vertical of the axis of the second polarizing filter is equal to 30.2°

4 0

3 years ago

Hans observed properties of four different waves and recorded observations about each one in his chart.

Andre45 [30]

D.Wave 3 resulted from constructive interference, and Wave 4 resulted from destructive interference.

7 0

3 years ago

Read 2 more answers

Two charges, one of 2.50μC and the other of -3.50μC, are placed on the x-axis, one at the origin and the other at x = 0.600 m

aev [14]

Answer:

Explanation:

Given

charge of first body $q_1=2.5\ mu C$

charge of second body $q_2=-3.5\ mu C$

Particle 1 is at origin and particle 2 is at $x=0.6\ m$

third Particle which charge +q must be placed left of $2.5\mu C$ because it will repel the q charge while $-3.5\mu C$ will attract it

suppose it is placed at a distance of x m

$F_{1q}=\frac{kq(2.5)}{x^2}$

$F_{2q}=\frac{kq(-3.5)}{(0.6+x)^2}$

$F_{1q}+F_{2q}=0$

$\frac{kq(2.5)}{x^2}+\frac{kq(-3.5)}{(0.6+x)^2}=0$

$\frac{kq(2.5)}{x^2}=\frac{kq(3.5)}{(0.6+x)^2}$

$\frac{0.6+x}{x}=(\frac{3.5}{2.5})^{0.5}$

$0.6+x=1.1832x$

$x=3.27\ m$

5 0

3 years ago

What is the is the 94th element of the periodic table

shutvik [7]

Answer:

plutonium

Explanation:

6 0

3 years ago

A 150 kg safe on frictionless casters is being raised at 1.20m to the bed of a truck using planks 4m long. The force needed to p

Vesnalui [34]

The only thing you need to know in order to solve this task is that plank length (which is force x), should equal the increase in potential energy, so what we have now : (mass)* g * (height).
It has to look like that: 
F * 3.0 = 150 x 9.81 x 1.20
Then solve for F, the result should be in newtones = 588N

Do hope it makes sense.

3 0

3 years ago