Unpolarized light of intensity is incident on a system of three polarizers: (i) The first polarizer has its axis vertical. (ii) The second polarizer is inclined at to the first. (iii) The third polarizer is horizontal (i.e., to the first). What is the final transmitted intensity?

Explanation: 1. Conceptual Framework To solve this problem, we need to understand two key principles of polarization: Transmission of Unpolarized Light: When completely unpolarized light passes through any ideal linear polarizer, its transmitted intensity drops exactly by half, regardless of the orientation of the transmission axis. Malus's Law: When completely plane-polarized light of intensity passes through a subsequent polarizer whose transmission axis makes an angle with the polarization plane of the incident light, the transmitted intensity is given by: 2. Step-by-Step Calculation Let's trace the light intensity through each of the three successive polarizers: Step A: Intensity after the First Polarizer ( ) The initial light is unpolarized with an intensity of . After passing through the first (v

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Explanation

1. Conceptual Framework

To solve this problem, we need to understand two key principles of polarization:

Transmission of Unpolarized Light: When completely unpolarized light passes through any ideal linear polarizer, its transmitted intensity drops exactly by half, regardless of the orientation of the transmission axis.
Malus's Law: When completely plane-polarized light of intensity $I_{in}$ passes through a subsequent polarizer whose transmission axis makes an angle $θ$ with the polarization plane of the incident light, the transmitted intensity $I_{out}$ is given by:
$I_{out} = I_{in} cos^{2} θ$

2. Step-by-Step Calculation

Let's trace the light intensity through each of the three successive polarizers:

Step A: Intensity after the First Polarizer ( $I_{1}$ )

The initial light is unpolarized with an intensity of $I_{0}$ . After passing through the first (vertical) polarizer, it becomes linearly polarized in the vertical direction, and its intensity becomes half of the initial value:

$I_{1} = \frac{I _{0}}{2}$

Step B: Intensity after the Second Polarizer ( $I_{2}$ )

The light incident on the second polarizer is now vertically polarized with intensity $I_{1}$ . The transmission axis of the second polarizer is inclined at an angle of $θ_{1} = 4 5^{\circ}$ relative to the first (vertical) axis.

Applying Malus's Law:

$I_{2} = I_{1} cos^{2} (4 5^{\circ})$

Since $cos (4 5^{\circ}) = \frac{1}{2}$ , we have:

$I_{2} = (\frac{I _{0}}{2}) \times (\frac{1}{2})^{2}$

$I_{2} = \frac{I _{0}}{2} \times \frac{1}{2} = \frac{I _{0}}{4}$

Now, the light emerging from the second polarizer is linearly polarized along its $4 5^{\circ}$ inclination angle.

Step C: Intensity after the Third Polarizer ( $I_{3}$ )

The light incident on the third polarizer has an intensity of $I_{2}$ and is polarized at a $4 5^{\circ}$ angle. The third polarizer is oriented horizontally ( $9 0^{\circ}$ to the first vertical polarizer).

The relative angle ( $θ_{2}$ ) between the second polarizer ( $4 5^{\circ}$ ) and the third polarizer ( $9 0^{\circ}$ ) is:

$θ_{2} = 9 0^{\circ} - 4 5^{\circ} = 4 5^{\circ}$

Applying Malus's Law once again for this stage:

$I_{3} = I_{2} cos^{2} (4 5^{\circ})$

$I_{3} = (\frac{I _{0}}{4}) \times (\frac{1}{2})^{2}$

$I_{3} = \frac{I _{0}}{4} \times \frac{1}{2} = \frac{I _{0}}{8}$

Conclusion

The final transmitted intensity after passing through the three-polarizer network is $\frac{I _{0}}{8}$ .

Correct Answer

The correct option is (d) $I_{0} /8$ .

Explanation

1. Conceptual Framework

To solve this problem, we need to understand two key principles of polarization:

Transmission of Unpolarized Light: When completely unpolarized light passes through any ideal linear polarizer, its transmitted intensity drops exactly by half, regardless of the orientation of the transmission axis.
Malus's Law: When completely plane-polarized light of intensity $I_{in}$ passes through a subsequent polarizer whose transmission axis makes an angle $θ$ with the polarization plane of the incident light, the transmitted intensity $I_{out}$ is given by:
$I_{out} = I_{in} cos^{2} θ$

2. Step-by-Step Calculation

Let's trace the light intensity through each of the three successive polarizers:

Step A: Intensity after the First Polarizer ( $I_{1}$ )

$I_{1} = \frac{I _{0}}{2}$

Step B: Intensity after the Second Polarizer ( $I_{2}$ )

Applying Malus's Law:

$I_{2} = I_{1} cos^{2} (4 5^{\circ})$

Since $cos (4 5^{\circ}) = \frac{1}{2}$ , we have:

$I_{2} = (\frac{I _{0}}{2}) \times (\frac{1}{2})^{2}$

$I_{2} = \frac{I _{0}}{2} \times \frac{1}{2} = \frac{I _{0}}{4}$

Now, the light emerging from the second polarizer is linearly polarized along its $4 5^{\circ}$ inclination angle.

Step C: Intensity after the Third Polarizer ( $I_{3}$ )

The relative angle ( $θ_{2}$ ) between the second polarizer ( $4 5^{\circ}$ ) and the third polarizer ( $9 0^{\circ}$ ) is:

$θ_{2} = 9 0^{\circ} - 4 5^{\circ} = 4 5^{\circ}$

Applying Malus's Law once again for this stage:

$I_{3} = I_{2} cos^{2} (4 5^{\circ})$

$I_{3} = (\frac{I _{0}}{4}) \times (\frac{1}{2})^{2}$

$I_{3} = \frac{I _{0}}{4} \times \frac{1}{2} = \frac{I _{0}}{8}$

Conclusion

The final transmitted intensity after passing through the three-polarizer network is $\frac{I _{0}}{8}$ .

Correct Answer

The correct option is (d) $I_{0} /8$ .

AMU 2026 — science PYQ

Choose the correct answer:

Explanation

1. Conceptual Framework

2. Step-by-Step Calculation

Conclusion

Correct Answer

Explanation

1. Conceptual Framework

2. Step-by-Step Calculation

Conclusion

Correct Answer