91影视

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

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:

$I鈭�E^2$

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

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

Intensity is directly proportional to the electric field square:

$I鈭�E^2$

Writethefraction of intensity received as,

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

Here,- final intensity

$I_0$- Initial intensity

$E_0$鈥� 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鈥檚 eyes is $\frac{I_f}{I_0}=0.16$.

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