Question

At a beach, the light is generally partially polarized due to reflections off sand and water. At a particular beach on a particular day near sundown, the horizontal component of the electric field vector is $\mathbf{\text{2.3}}$times the vertical component. A standing sunbather puts on polarizing sunglasses; the glasses eliminate the horizontal field component.

(a) What fraction of the light intensity received before the glasses were put on now reaches the sunbather’s eyes?

(b) The sunbather, still wearing the glasses, lies on his side. What fraction of the light intensity received before the glasses were put on now reaches his eyes?

Short answer

1. The fraction of light intensity received before the glasses were put on that now reaches the sunbather’s eyes is $\frac{I_f}{I_0}=0.16$ .
2. After the sunbather still wearing the glasses lies on his side, the fraction of light intensity received before the glasses were put on that now reaches his eyes is .

Step-by-step solution

Step 1: Given

The horizontal component of the electric field is 2.3 times vertical component.

Determining the concept

Use the concept of the intensity of light related to the electric field. Using the components of the electric field, find fraction of light received by the sunbather. Electric field intensity is equal to the negative of rate of change of potential with respect to the distance or it can be defined as the negative of the rate of derivative of potential difference, V with respect tor , $\mathbf{E}\mathbf{=}\mathbf{-}\frac{\mathrm{dV}}{\mathrm{dr}}$.

Formulae are as follows:

$I\propto E^2$

Here, l is the intensity, and E is the electric field.

(a) Determine the fraction of light intensity received before the glasses were put on now reaches the sunbather’s eyes

Fraction of light intensity received before the glasses were put on that now reaches the sunbather’s eyes:

Intensity is directly proportional to the electric field square:

$I\propto E^2$

Writethefraction of intensity received as,

$\frac{I_f}{I_0}=\frac{E_f^2}{E_0^2}$ ……. (1)

Here,- final intensity

$I_0$- Initial intensity

$E_0$– Initial electric field

$E_f$- Final electric field

Theelectric field isthesum of horizontal and vertical components.

$E_0=E_x+E_y$

Using this, write equation (1) as,

$$\begin{gathered} \frac{I_f}{I_0}=\frac{E_y^2}{E_x^2+E_y^2} \\ {E}_x=2.3E_y, \\ \frac{I_f}{I_0}=\frac{E_y^2}{{\left(2.3E_y\right)}^2+E_y^2} \end{gathered}$$

Substitute the values and solve as:

$$\begin{gathered} \frac{I_f}{I_0}=\frac{1}{{\left(2.3\right)}^2+1} \\ \frac{I_f}{I_0}=0.16 \end{gathered}$$

The fraction of light intensity received before the glasses were put on that now reaches the sunbather’s eyes is $\frac{I_f}{I_0}=0.16$.

(b) Determine the fraction of light intensity received before the glasses were put on now reaches his eyes when the sunbather, still wearing the glasses, lies on his side

After the sunbather still wearing the glasses lies on his side, the fraction of light intensity received before the glasses were put on that now reaches his eyes:

In this case, the horizontal component of the electric field passes through the glasses.

$$\begin{gathered} \frac{I_f}{I_0}=\frac{E_x^2}{E_x^2+E_y^2} \\ {E}_x=2.3E_y \end{gathered}$$

$$\begin{gathered} \frac{I_f}{I_0}=\frac{{\left(2.3E_y\right)}^2}{{\left(2.3E_y\right)}^2+E_y^2} \\ \frac{I_f}{I_0}=\frac{{2.3}^2}{{2.3}^2+1} \end{gathered}$$

Solve further as:

$\frac{I_f}{I_0}=0.84$

The intensity of light is directly proportional to the square of the electric field. Using this concept, solve the above problem