Question

The intensity of transmitted light when a polaroid sheet, placed between two crossed polaroids at 22.5° from the polarization axis of one of the polaroid, is ($I_0$ is the intensity of polarised light after passing through the first polaroid):

• A. $\frac{I_0}{8}$
• B. $\frac{I_0}{16}$
• C. $\frac{I_0}{2}$
• D. $\frac{I_0}{4}$

Answer 1

Answer: (A)

$\frac{I_0}{8}$

Answer 2

The problem describes a system of three polaroids. Let the first polaroid be P1, the second polaroid be P3 (the one in the middle), and the third polaroid be P2.

P1 and P2 are crossed polaroids, meaning their polarization axes are perpendicular to each other. Let the polarization axis of P1 be along the x-direction (angle 0°). Since P1 and P2 are crossed, the polarization axis of P2 is along the y-direction (angle 90°). The third polaroid P3 is placed between P1 and P2, with its polarization axis at an angle of 22.5° from the polarization axis of P1. So, the angle between the axis of P1 and P3 is $\theta = 22.5^\circ$.

The intensity of polarized light after passing through the first polaroid (P1) is given as $I_0$. This light is polarized along the axis of P1.

This polarized light of intensity $I_0$, polarized along the axis of P1, is incident on the polaroid P3. The angle between the polarization direction of the incident light (axis of P1) and the transmission axis of P3 is $\theta = 22.5^\circ$. According to Malus's Law, the intensity of light transmitted through P3, let's call it $I_1$, is given by:

$I_1 = I_0 \cos^2 \theta = I_0 \cos^2(22.5^\circ)$.

The light transmitted through P3 is polarized along the axis of P3.

This light of intensity $I_1$, polarized along the axis of P3 (at angle 22.5° relative to P1), is then incident on the polaroid P2. The transmission axis of P2 is at an angle of 90° relative to the axis of P1. The angle between the polarization direction of the light incident on P2 (axis of P3, at 22.5°) and the transmission axis of P2 (at 90°) is $\phi = 90^\circ - 22.5^\circ = 67.5^\circ$. According to Malus's Law, the intensity of light transmitted through P2, which is the final transmitted intensity $I_{transmitted}$, is given by:

$I_{transmitted} = I_1 \cos^2 \phi = I_1 \cos^2(67.5^\circ)$.

Substitute the expression for $I_1$:

$I_{transmitted} = (I_0 \cos^2(22.5^\circ)) \cos^2(67.5^\circ) = I_0 \cos^2(22.5^\circ) \cos^2(67.5^\circ)$.

We can use the trigonometric identity $\cos(90^\circ - \alpha) = \sin(\alpha)$. So, $\cos(67.5^\circ) = \cos(90^\circ - 22.5^\circ) = \sin(22.5^\circ)$.

Thus, $I_{transmitted} = I_0 \cos^2(22.5^\circ) \sin^2(22.5^\circ)$.

We can use the identity $\sin(2\alpha) = 2 \sin\alpha \cos\alpha$, which means $\sin\alpha \cos\alpha = \frac{1}{2} \sin(2\alpha)$. Squaring this, we get $\sin^2\alpha \cos^2\alpha = \left(\frac{1}{2} \sin(2\alpha)\right)^2 = \frac{1}{4} \sin^2(2\alpha)$. Let $\alpha = 22.5^\circ$. Then $2\alpha = 45^\circ$.

$I_{transmitted} = I_0 \left(\frac{1}{4} \sin^2(2 \times 22.5^\circ)\right) = I_0 \left(\frac{1}{4} \sin^2(45^\circ)\right)$.

We know that $\sin(45^\circ) = \frac{1}{\sqrt{2}}$. So, $\sin^2(45^\circ) = \left(\frac{1}{\sqrt{2}}\right)^2 = \frac{1}{2}$.

Substituting this value:

$I_{transmitted} = I_0 \left(\frac{1}{4} \times \frac{1}{2}\right) = I_0 \left(\frac{1}{8}\right) = \frac{I_0}{8}$.

The intensity of the transmitted light is $\frac{I_0}{8}$.