Answer:
A) I_f3 = 27.58 W/cm²
B) I_f2 = 102.26 W/cm²
Step-by-step 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