6.Two polarizers are oriented at 48^circ to each other and plane -polarized light is incident on them.If only 35% of the light gets through both of them, what was the initial polarization direction of the incident light ()

To find the initial polarization direction of the incident light, we need to consider the Malus's law, which relates the intensity of light after passing through a polarizer to the angle between the polarization direction of the incident light and the axis of the polarizer.<br /><br />Malus's law is given by:<br /><br />\[ I = I_0 \cdot \cos^2(\theta) \]<br /><br />where:<br />- \( I \) is the intensity of the light after passing through the polarizer,<br />- \( I_0 \) is the initial intensity of the light,<br />- \( \theta \) is the angle between the polarization direction of the incident light and the axis of the polarizer.<br /><br />In this case, we are given that only 35% of the light gets through both polarizers. This means that the intensity of the light after passing through both polarizers is 35% of the initial intensity.<br /><br />Let's denote the initial polarization direction of the incident light as \( \alpha \). The light first passes through a polarizer oriented at an angle of \( \alpha \) to the incident polarization direction, and then through a second polarizer oriented at an angle of \( 48^\circ \) to the first polarizer.<br /><br />Using Malus's law for the first polarizer, we have:<br /><br />\[ I_1 = I_0 \cdot \cos^2(\alpha) \]<br /><br />Using Malus's law for the second polarizer, we have:<br /><br />\[ I = I_1 \cdot \cos^2(48^\circ) \]<br /><br />Substituting the expression for \( I_1 \) from the first equation into the second equation, we get:<br /><br />\[ I = I_0 \cdot \cos^2(\alpha) \cdot \cos^2(48^\circ) \]<br /><br />We are given that \( I = 0.35 \cdot I_0 \), so we can substitute this into the equation:<br /><br />\[ 0.35 \cdot I_0 = I_0 \cdot \cos^2(\alpha) \cdot \cos^2(48^\circ) \]<br /><br />Dividing both sides by \( I_0 \), we get:<br /><br />\[ 0.35 = \cos^2(\alpha) \cdot \cos^2(48^\circ) \]<br /><br />To find the value of \( \alpha \), we need to solve this equation. Taking the square root of both sides, we get:<br /><br />\[ \sqrt{0.35} = \cos(\alpha) \cdot \cos(48^\circ) \]<br /><br />Dividing both sides by \( \cos(48^\circ) \), we get:<br /><br />\[ \frac{\sqrt{0.35}}{\cos(48^\circ)} = \cos(\alpha) \]<br /><br />Finally, taking the inverse cosine of both sides, we get:<br /><br />\[ \alpha = \cos^{-1}\left(\frac{\sqrt{0.35}}{\cos(48^\circ)}\right) \]<br /><br />Calculating this value, we find that \( \alpha \) is approximately \( 55.56^\circ \). Therefore, the initial polarization direction of the incident light was \( 55.56^\circ \) relative to the axis of the first polarizer.