Chapter 24, Problem 53P

To determine

The fraction of the light intensity that is transmitted

Answer to Problem 53P

  $0.089$ .

Explanation of Solution

Given:

Angle of reflection, $\theta =65°$

Formula used:

According to Malus law,

If $\theta$ be the angle between the direction of electric field and polarization axis and $I_0$ be the intensity light. So, the intensity of transmitted: $\text{I}={\text{I}}_0{\cos }^2\text{θ}$

Calculation:

Let the initial intensity is ${\text{I}}_0$ through the two sheets.

According to the Malus law,

  ${\text{I}}_{\text{2}}\text{=}{\text{I}}_{\text{1}}{\text{cos}}^{\text{2}}\text{θ}\text{=}\frac{\text{1}}{\text{2}}{\text{I}}_{\text{0}}{\text{cos}}^{\text{2}}\text{θ}$

Hence,

  $\frac{{\text{I}}_{\text{2}}}{{\text{I}}_{\text{0}}}\text{=}\frac{{\text{cos}}^{\text{2}}\text{θ}}{\text{2}}\text{=}\frac{{\text{cos}}^{\text{2}}\text{65°}}{\text{2}}\text{=}\text{0}\text{.089}$

Conclusion:

The fraction of the light intensity that is transmitted is $0.089$ .