Question

Unpolarized light whose intensity is 1.46 W/m2 is incident on the polarizer in the drawing. (a) What is the intensity of the light leaving the polarizer? (b) If the analyzer is set at an angle of = 54.0° with respect to the polarizer, what is the intensity of the light that reaches the photocell?

Answer 1

Answer:

0.73 W/m²

0.2522 W/m²

Explanation:

$I_0$ = Unpolarized light = 1.46 W/m²

$\theta$ = Analyzer angle = 54°

Light through first filter

$I_1=\frac{I_0}{2}\\\Rightarrow I_1=\frac{1.46}{2}\\\Rightarrow I_1=0.73\ W/m^2$

The intensity of the light leaving the polarizer is 0.73 W/m²

After passing through analyzer

$I=I_1cos^2\theta\\\Rightarrow I=0.73\times cos^2(54)\\\Rightarrow I=0.2522\ W/m^2$

The intensity of the light that reaches the photocell is 0.2522 W/m²

Answer 2

The intensity of the light leaving the polarizer and  intensity of the light that reaches the photocell is mathematically given as

I1=0.73 W/m^2

I=0.2522 W/m^2

What is  the intensity of the light leaving the polarizer and  intensity of the light that reaches the photocell?

Question Parameter(s):

Unpolarized light whose intensity is 1.46 W/m2

at an angle of = 54.0°

Generally, the equation for the Intensity  is mathematically given as

I_1=I_0/2

Therefore

I1=1.46/2
I1=0.73 W/m^2

In conclusion,  as it passes through analyzer

I=0.73* cos^2(54)

I=0.2522 W/m^2

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