Question

Unpolarized light with intensity 50.0 W m^-2 passes first through a polarizing filter with its transmission axis horizontal, and then through a second filter. The transmitted light intensity is 18.0 W m^-2 . The transmission axis of the second filter is at:

(a). 31.9o from the horizontal

(b). 43.9o from the horizontal

(c). 84.9o from the horizontal

(d). 43.9o from the vertical

(e). 31.9o from the vertical

I am confused about the transmission axis in this question. The formula is I=I0*cos^2(theta). So, is this theta equals to the angle between light I0 and the filter surface, or the normal line of filter suface?

Answer 1

I_0 = 50 w/m^2 after first horizontal I_1 = I_0/2 = 50/2 = 25 w/m^2 I_2 = I_1 cos^2 theta 1 theta = 25 c0s^2 theta theta = 31.9 degree from vertical