1. Given data

• Initial unpolarised intensity: $I_0 = 32\,\text{Wm}^{-2}$
• Three polaroids are used.
• Axis of the 3rd polaroid is perpendicular to axis of the 1st polaroid.
• Final emergent intensity: $I = 3\,\text{Wm}^{-2}$

Let the angle between the pass axes of the first and second polaroids be $\theta$.

Since the axis of the third is perpendicular to the first, the angle between the second and third polaroids is: $90^\circ - \theta$

2. Intensity after first polaroid

For unpolarised light passing through a polaroid: $I_1 = \frac{I_0}{2} = \frac{32}{2} = 16\,\text{Wm}^{-2}$

3. Intensity after second polaroid

By Malus' law: $I_2 = I_1 \cos^2\theta = 16\cos^2\theta$

4. Intensity after third polaroid

The angle between second and third axes is $90^\circ-\theta$, so again by Malus' law: $I = I_2 \cos^2(90^\circ-\theta)$ But, $\cos(90^\circ-\theta)=\sin\theta$ So, $I = 16\cos^2\theta\sin^2\theta$

Given $I=3$, hence $16\cos^2\theta\sin^2\theta = 3$

5. Simplify the trigonometric expression

Using $\sin^2 2\theta = 4\sin^2\theta\cos^2\theta$ we get $16\cos^2\theta\sin^2\theta = 16\cdot \frac{\sin^2 2\theta}{4} = 4\sin^2 2\theta$

So, $4\sin^2 2\theta = 3$ $\sin^2 2\theta = \frac{3}{4}$ $\sin 2\theta = \pm \frac{\sqrt{3}}{2}$

Thus, $2\theta = 60^\circ, 120^\circ$ which gives $\theta = 30^\circ \quad \text{or} \quad 60^\circ$

6. Final answer

The angle between the pass axes of the first two polaroids can be: $\boxed{30^\circ \text{ or } 60^\circ}$