A point source $$S$$ emits unpolarized light uniformly in all directions. At two points $$A$$ and $$B$$, the ratio $$r = I_A/I_B$$ of the intensities of light is 2. If a set of two polaroids having 45° angle between their pass-axes is placed just before point $$B$$, then the new value of $$r$$ will be ______.

Let $$I_A$$ and $$I_B$$ be the intensities at points $$A$$ and $$B$$ before inserting the polaroids.
Given $$\dfrac{I_A}{I_B}=2 \; \Rightarrow \; I_A = 2\,I_B$$ $$-(1)$$

Now place two polaroids in front of point $$B$$. Call the initial unpolarized intensity reaching the first polaroid $$I_B$$.

Step 1: First polaroid
Transmission of unpolarized light through a single polaroid is $$\dfrac{1}{2}$$ of the incident intensity.
Therefore the intensity emerging from the first polaroid is
$$I_1 = \dfrac{I_B}{2}$$ $$-(2)$$
The light is now plane-polarized along the pass-axis of the first polaroid.

Step 2: Second polaroid
The pass-axis of the second polaroid is at $$45^{\circ}$$ to that of the first. For plane-polarized light, Malus’ law states
$$I = I_{\text{in}} \cos^{2}\theta$$
where $$\theta$$ is the angle between the light’s polarization direction and the pass-axis of the analyzer (second polaroid). Here $$\theta = 45^{\circ}$$, so $$\cos^{2}45^{\circ} = \left(\dfrac{1}{\sqrt 2}\right)^2 = \dfrac{1}{2}$$.

Hence the intensity after the second polaroid is
$$I_B' = I_1 \cos^{2}45^{\circ} = \dfrac{I_B}{2} \times \dfrac{1}{2} = \dfrac{I_B}{4}$$ $$-(3)$$

Step 3: New intensity ratio
Point $$A$$ is unaffected, so its intensity remains $$I_A$$. The new ratio of intensities is
$$r_{\text{new}} = \dfrac{I_A}{I_B'} = \dfrac{I_A}{\dfrac{I_B}{4}} = 4 \dfrac{I_A}{I_B}$$

Using $$(1)$$, $$\dfrac{I_A}{I_B}=2$$, therefore
$$r_{\text{new}} = 4 \times 2 = 8$$

Hence the new value of $$r$$ is $$8$$.