Intensity Ratio of Point Source Light through Polaroids

To find the new ratio of light intensities at points $A$ and $B$, we analyze the effect of placing two polaroids in front of point $B$.

1. Initial State: The point source $S$ emits unpolarized light. Let $I_A$ and $I_B$ be the initial intensities at points $A$ and $B$, respectively. The initial ratio of intensities is given by: $r = \frac{I_A}{I_B} = 2$

2. Intensity after passing through the first polaroid: When unpolarized light of intensity $I_B$ falls on the first polaroid, the intensity of the transmitted linearly polarized light, $I_1$, becomes half of the incident intensity: $I_1 = \frac{I_B}{2}$

3. Intensity after passing through the second polaroid: The second polaroid has its pass-axis oriented at an angle of $\theta = 45^\circ$ with respect to the pass-axis of the first polaroid. According to Malus's Law, the new intensity $I_B'$ transmitted through the second polaroid just before point $B$ is: $I_B' = I_1 \cos^2(45^\circ)$

   Substituting $I_1 = \frac{I_B}{2}$ and $\cos(45^\circ) = \frac{1}{\sqrt{2}}$: $I_B' = \left(\frac{I_B}{2}\right) \left(\frac{1}{\sqrt{2}}\right)^2 = \left(\frac{I_B}{2}\right) \left(\frac{1}{2}\right) = \frac{I_B}{4}$

4. New Ratio of Intensities: The intensity at point $A$ remains unchanged, i.e., $I_A$. The new value of the ratio $r_{\text{new}}$ is: $r_{\text{new}} = \frac{I_A}{I_B'} = \frac{I_A}{\frac{I_B}{4}} = 4 \left(\frac{I_A}{I_B}\right)$

   Substituting the given initial ratio $\frac{I_A}{I_B} = 2$: $r_{\text{new}} = 4 \times 2 = 8$

Thus, the new value of $r$ is 8.