Young's Double Slit Polarizer Effect

This problem calculates the visibility of an interference pattern in a Young's double-slit experiment where a linear polarizer affects the light intensity and polarization for one slit.

Polarizer Intensity Reduction Calculation

The incident light is monochromatic and linearly polarized. Let its initial intensity be $I_0$ and its polarization direction be $\hat{p}_1$.

• A linear polarizer is placed in front of one slit (slit 2). Its pass axis ($\hat{a}$) is at an angle of $45^\circ$ to the incident polarization $\hat{p}_1$.
• The intensity of light reaching slit 1 remains $I_1 = I_0$. Its polarization is $\hat{p}_1$.
• According to Malus's Law, the intensity of light transmitted through the polarizer to slit 2 is reduced: $I_2 = I_0 \cos^2(45^\circ) = I_0 \left(\frac{1}{\sqrt{2}}\right)^2 = \frac{1}{2} I_0$.
• The polarization direction of the light reaching slit 2 is along the polarizer's pass axis, $\hat{p}_2 = \hat{a}$.

Visibility Formula Application

The visibility ($V_{std}$) is fundamentally related to the intensities ($I_1, I_2$) of the waves contributing to the interference:

$V_{std} = \frac{I_{max} - I_{min}}{I_{max} + I_{min}} = \frac{2 \sqrt{I_1 I_2}}{I_1 + I_2}$

Substitute the intensities $I_1 = I_0$ and $I_2 = \frac{1}{2} I_0$ into the formula:

$V_{std} = \frac{2 \sqrt{I_0 \cdot \frac{1}{2} I_0}}{I_0 + \frac{1}{2} I_0} = \frac{2 \sqrt{\frac{1}{2} I_0^2}}{\frac{3}{2} I_0} = \frac{2 \cdot \frac{1}{\sqrt{2}} I_0}{\frac{3}{2} I_0}$

Simplifying this expression gives:

$V_{std} = \frac{\sqrt{2} I_0}{\frac{3}{2} I_0} = \frac{2\sqrt{2}}{3}$

Polarization Mismatch Impact

The polarizer alters the polarization state. The light from slit 1 has polarization $\hat{p}_1$, and from slit 2 has polarization $\hat{p}_2$, with an angle of $45^\circ$ between them. The factor representing this mismatch is $|\hat{p}_1 \cdot \hat{p}_2| = \cos(45^\circ) = \frac{1}{\sqrt{2}}$.

The effective visibility ($V$) accounts for this polarization difference:

$V = V_{std} |\hat{p}_1 \cdot \hat{p}_2|$

Substituting the calculated $V_{std}$ and the polarization mismatch factor:

$V = \frac{2\sqrt{2}}{3} \cdot \frac{1}{\sqrt{2}} = \frac{2}{3}$

Final Visibility Value

The resulting visibility of the interference pattern is $\frac{2}{3}$.