The diagram shows combination of polaroids.Unpolarised light of intensity I₀ incident perpendicular to the axis of polaroid P₁, then angle θ for which maximum intensity passes through polaroid P₃

Correct Answer: 45°

Step-by-Step Explanation:

1. Understanding the Setup from the Diagram:
Let us analyze the alignment of the transmission axes of the three polaroids shown in the image:
- The first polaroid, P₁, has its transmission axis oriented vertically.
- The second polaroid, P₂, has its transmission axis oriented at an angle θ with respect to the vertical axis of P₁.
- The third polaroid, P₃, has its transmission axis oriented horizontally, which is perpendicular to the vertical axis of P₁.

2. Applying Malus's Law at each Polarisation stage:

Step 1: Transmission through P₁
Initially, unpolarised light of intensity I₀ is incident on the first polaroid P₁. When unpolarised light passes through a polaroid, its intensity is reduced to exactly half, and it becomes linearly polarised along the transmission axis of the polaroid. Therefore, the intensity I₁ after passing through P₁ is:

$I_1=\frac{I_0}{2}$

Step 2: Transmission through P₂
The light incident on P₂ is now polarised vertically. Since the axis of P₂ is at an angle θ to the vertical axis of P₁, we use Malus's Law to calculate the transmitted intensity I₂:

$I_2=I_1{\cos }^2\theta =\frac{I_0}{2}{\cos }^2\theta$

Step 3: Transmission through P₃
The light passing out of P₂ is polarised along the transmission axis of P₂. This light is then incident on the third polaroid P₃. Since the axis of P₁ is vertical and the axis of P₃ is horizontal, the angle between the vertical and the axis of P₃ is 90°. The axis of P₂ is inclined at an angle θ to the vertical, so the angle between the transmission axes of P₂ and P₃ is (90° - θ).
Applying Malus's Law once more, the final transmitted intensity I₃ is:

$I_3=I_2{\cos }^2(90°-\theta )=I_2{\sin }^2\theta$

Substituting the expression for I₂ into the equation above:

$I_3=\frac{I_0}{2}{\cos }^2\theta {\sin }^2\theta$

Using the trigonometric double-angle identity, we can simplify this expression:
$\sin \theta \cos \theta =\frac{\sin 2\theta }{2}$
Squaring both sides gives:
${\sin }^2\theta {\cos }^2\theta =\frac{{\sin }^{2}2\theta }{4}$

Now, substitute this back into the formula for I₃:

$I_3=\frac{I_0}{2}\times \frac{{\sin }^{2}2\theta }{4}=\frac{I_0}{8}{\sin }^{2}2\theta$

3. Finding the condition for Maximum Intensity:
The intensity of the light emerging from P₃ is maximum when the value of the sine term is at its maximum value:

${\sin }^{2}2\theta =1$

Taking the positive square root for acute angles:

$\sin 2\theta =1$

$2\theta =90°$

$\theta =45°$

Thus, the angle θ for which the maximum intensity of light passes through polaroid P₃ is 45°.