Three ideal polarizing filters are stacked, with the polarizing axis of the second and third filters at 21 degrees and 61 degrees, respectively, to that of the first. If unpolarized light is incident on the stack, the light has intensity 60.0 w/cm
2
after it passes through the stack.

If the incident intensity is kept constant:

1) What is the intensity of the light after it has passed through the stack if the second polarizer is removed?

2) What is the intensity of the light after it has passed through the stack if the third polarizer is removed?

A) I_f3 = 27.58 W/cm²

B) I_f2 = 102.26 W/cm²

Explanation:

We are given;

-The angle of the second polarizing to the first; θ_2 = 21°

-The angle of the third  polarizing to the first; θ_1 = 61°

- The unpolarized light after it pass through the polarizing stack; I_u = I_3 = 60 W/cm² 

A) Let the initial intensity of the beam of light before polarization be I_p

Thus, when the unpolarized light passes through the first polarizing filter, the intensity of light that emerges would be given as;

I_1 = (I_p)/2

According to Malus’s law,

I = I_max(cos²Φ)

Thus, we can say that;

the intensity of light that would emerge from the second polarizing filter would be given as;

I_2 = I_1(cos²Φ1) = ((I_p)/2)(cos²Φ1)

Similarly, the intensity of light that will emerge from the third filter would be given as;

I_3 = I_2(cos²Φ1) = ((I_p)/2)(cos²Φ1)(cos²(Φ2 - Φ1)

Thus, making I_p the subject of the formula, we have ;

I_p = (2I_3)/[(cos²Φ1)(cos²(Φ2 - Φ1)]

Plugging in the relevant values, we have;

I_p = (2*60)/[(cos²21)(cos²(61 - 21)]

I_p = 234.65 W/cm²

Now, when the second polarizer is removed, the third polarizer becomes the second and final polarizer so the intensity of light emerging from the stack would be given as;

I_f3 = (I_p/2)(cos²Φ2)

I_f3 = (234.65/2)(cos²61)

I_f3 = 27.58 W/cm²

B) Similarly, when the third polarizer is removed, the second polarizer becomes the final polarizer and the intensity of light emerging from the stack would be given as;

I_f2 = (I_p/2)(cos²Φ1)

I_f2 = (234.65/2)(cos²21)

I_f2 = 102.26 W/cm²

f=ma a/f/m a= 480/60 a = 8m/s/s that is quick wow

  distance on the y-axis and time on the x-axis.

explanation:

the instantaneous velocity = change in position/time taken

the slope of the graph should give instantaneous velocity.

  slope of a graph = change in value of y -axis / change in values of x -axis

  comparing both the equations

    change in value of y -axis = change in position

  change in values of x -axis = time taken

so position values should be on the y axis and time values should be on the x axis.

distance on the y-axis and time on the x-axis.