Question

In Fig. 34-42, initially unpolar ized light is sent through three polarizing sheets whose polarizing directions make angles of 1 = 44.2°, 2 = 20.3°, and 3 = 41.2° with the direction of the y axis. What percentage of the light's initial intensity is transmitted by the system? (Hint: Be careful with the angles.) %

the angles.) Fig. 34-42

Answer 1

When a light passes through a polariser, then the intensity of the light which comes out of the polarizer is:

$I_{p} = I_{o}\times cos(\theta )^{2}$

Where: $\theta$ is the angle between incoming polarized light and the axis of polarizer

Lets say the initial inensity of the light is : $I_{0}$

Step1) When the light passes through first polarizing Sheet then its intensity wil become $I_{p1}$ and will be polarized along that angle.

$I_{p1} = I_{o}\times cos(\theta )^{2}$

Note: $\theta$ = 44.2 deg = 0.77 radians

$I_{p1} = I_{o}\times cos(0.77 )^{2} = 0.5I_{o}$

But now this light is polarised at an angle of 44.2 deg from the y-direction

Therefore, in the next case: $\theta$ wil be: 180 - 44.2-20.3 = 115.5 deg = 2 radians

and final intensity after passing through the second polarising sheet is:

$I_{p2} = I_{p1}\times cos(\theta )^{2}$

$I_{p2} = 0.5I_{o}\times cos(2)^{2}$

Therefore, in the next case: $\theta$ wil be: 180 - 20.3 - 41.2 = 118.5 deg = 2.07 radians

and final intensity after passing through the second polarising sheet is:

$I_{p3} = I_{p2}\times cos(\theta )^{2}$

$I_{p3} = 0.086I_{o}\times cos(2.07)^{2}$

Percentage of light transmitted:

$Perc = \frac{0.019I_{o}}{I_{o}}\times 100$

$Perc = \frac{0.019}{1}\times 100$

Happy Learning :D

Answer 2

When a light passes through a polariser, then the intensity of the light which comes out of the polarizer is:

$I_{p} = I_{o}\times cos(\theta )^{2}$

Where: $\theta$ is the angle between incoming polarized light and the axis of polarizer

Lets say the initial inensity of the light is : $I_{0}$

Step1) When the light passes through first polarizing Sheet then its intensity wil become $I_{p1}$ and will be polarized along that angle.

$I_{p1} = I_{o}\times cos(\theta )^{2}$

Note: $\theta$ = 44.2 deg = 0.77 radians

$I_{p1} = I_{o}\times cos(0.77 )^{2} = 0.5I_{o}$

But now this light is polarised at an angle of 44.2 deg from the y-direction

Therefore, in the next case: $\theta$ wil be: 180 - 44.2-20.3 = 115.5 deg = 2 radians

and final intensity after passing through the second polarising sheet is:

$I_{p2} = I_{p1}\times cos(\theta )^{2}$

$I_{p2} = 0.5I_{o}\times cos(2)^{2}$

Therefore, in the next case: $\theta$ wil be: 180 - 20.3 - 41.2 = 118.5 deg = 2.07 radians

and final intensity after passing through the second polarising sheet is:

$I_{p3} = I_{p2}\times cos(\theta )^{2}$

$I_{p3} = 0.086I_{o}\times cos(2.07)^{2}$

Percentage of light transmitted:

$Perc = \frac{0.019I_{o}}{I_{o}}\times 100$

$Perc = \frac{0.019}{1}\times 100$

Happy Learning :D