Question

unpolarized light with intensity 400 w/m2 passes first through a polarizing filter with its axis vertical, then through a second polarizing filter. it emerges from the second filter with intensity 141 w/m2 . part a what is the angle from vertical of the axis of the second polarizing filter? express your answer with the appropriate units.

Answer 1

Approximately
`32.9^(\circ)`.

Step-by-step explanation:

When unpolarized light goes through a polarizing filter, intensity of the light would be reduced to
`(1/2)` of the initial value. In this case, intensity of the light would be reduced to
`200\; {\rm W\cdot m^(-2)}` after entering the first filter.

Malus's Law models the intensity of the light after going through the second filter:

`I_(1) = I_(0)\, \left(\cos(\theta)\right)^(2)`,

Where:

• 
  `I_(0) = 200\; {\rm W\cdot m^(-2)}}` is the intensity of the light before entering this polarizing filter.
• 
  `I_(1) = 141\; {\rm W\cdot m^(-2)}` is the intensity of the light after going through this filter.
• 
  `\theta` is the angle between the vertical axis of the filter and the plane of the incoming light.

Note that in this question, after entering the first polarizing filter, the plane of light would be parallel to the vertical axis of the first filter. Hence, the angle
`\theta` would also be equal to the angle between the vertical axes of the two filters.

Rearrange this equation to find
`\theta`:

`\displaystyle (\cos(\theta))^(2) = (I_(1))/(I_(0))`.

`\begin{aligned} \theta &= \arccos \sqrt{(I_(1))/(I_(0))} \\ &= \arccos \sqrt{(141)/(200)} \\ &\approx 32.9^(\circ)\end{aligned}`.