Question

Polarized light passes through a polarizing filter which has an axis of transmission oriented at θ₁ = 39.4 degrees from vertical. The intensity of the emerging light is measured to be I1 = 700 W/m2. The polarizer is then rotated so that its axis of transmission is θ₂ = 70.9 degrees from vertical, and the emerging light is now measured to be 290 W/m . Find the intensity of the incident polarized light and its orientation with respect to vertical. Answer must be 812 W/m², 17.6 degrees

Answer 1

angle from vertical =

intensity = I_o

case 1 :

I₁ = 700

700 = I_o Cos² (39.4 - )                      eq-1

case 2 :

I₁ = 290

290 = I_o Cos² (70.9 - )                      eq-2

dividing eq-1 by eq-2

700 / 290 = Cos² (39.4 - ) /Cos² (70.9 - )

sqrt(700/290) = (cos39.4 Cos + Sin39.4 Sin) / (cos70.9 Cos + Sin70.9 Sin)

1.6 = (0.77 Cos + 0.63 sqrt(1 - cos²) ) / (0.33 Cos + 0.94 sqrt(1 - cos²) )

= 17.6

using eq-1

700 = I_o Cos² (39.4 - 17.6)  

I_o = 812