Malus Law Quiz: Understanding Polarized Light Concepts

1. Malus’ law relates transmitted intensity through a polarizer to:

Wavelength only

The angle between polarization direction and the polarizer axis

The mass of the filter

The speed of light in vacuum

Malus’ law describes how the transmitted intensity depends on orientation. The key idea is that only the component of the electric field along the axis passes.

Explanation

Malus’ law describes how the transmitted intensity depends on orientation. The key idea is that only the component of the electric field along the axis passes.

2. If polarized light is aligned with the polarizer axis, transmission is maximum.

True

False

Maximum transmission at 0°. When the polarization direction matches the axis, the full field component passes. This gives the largest transmitted intensity for that incoming beam.

Explanation

Maximum transmission at 0°. When the polarization direction matches the axis, the full field component passes. This gives the largest transmitted intensity for that incoming beam.

3. If two ideal polarizers are crossed at 90°, the transmitted intensity (for initially polarized light aligned to the first) is:

Maximum

Zero

Half

Double

The first polarizer sets a polarization direction. The second at 90° blocks that direction, giving (ideally) no transmitted light.

Explanation

The first polarizer sets a polarization direction. The second at 90° blocks that direction, giving (ideally) no transmitted light.

4. For unpolarized light, the first ideal polarizer transmits about ______ of the intensity.

Unpolarized light has no preferred direction, so on average half the field component aligns with the axis. That yields ~50% intensity transmission.

Explanation

Unpolarized light has no preferred direction, so on average half the field component aligns with the axis. That yields ~50% intensity transmission.

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5. Malus’ law is commonly written as:

(i = i_0 \sin\theta)

(i = i_0 \cos^2\theta)

(i = i_0 \theta^2)

(i = i_0 / \cos\theta)

The transmitted intensity depends on the squared cosine of the angle. This comes from projecting the electric field onto the transmission axis.

Explanation

The transmitted intensity depends on the squared cosine of the angle. This comes from projecting the electric field onto the transmission axis.

6. If (\theta = 60^\circ), then (\cos^2\theta =):

1

1/2

1/4

3/4

(\cos 60^\circ = 1/2). Squaring gives ((1/2)^2 = 1/4).

Explanation

(\cos 60^\circ = 1/2). Squaring gives ((1/2)^2 = 1/4).

7. Polarized light of intensity 8 units passes a polarizer at 60°. Output intensity is:

2 units

4 units

8 units

Use (i = i_0 \cos^2\theta). With (i_0=8) and (\cos^2 60^\circ = 1/4), (i=2).

Explanation

Use (i = i_0 \cos^2\theta). With (i_0=8) and (\cos^2 60^\circ = 1/4), (i=2).

8. At (\theta = 45^\circ), Malus’ law predicts half the intensity passes for initially polarized light.

True

False

(\cos 45^\circ = \sqrt{2}/2), and squaring gives 1/2. So (i = i_0/2).

Explanation

(\cos 45^\circ = \sqrt{2}/2), and squaring gives 1/2. So (i = i_0/2).

9. If polarized light passes through a polarizer at 0°, then:

(i=0)

(i=i_0)

(i=i_0/2)

(i=2i_0)

(\cos 0^\circ = 1), so (\cos^2 0^\circ = 1). That means the polarizer passes the full intensity of that polarized component.

Explanation

(\cos 0^\circ = 1), so (\cos^2 0^\circ = 1). That means the polarizer passes the full intensity of that polarized component.

10. Malus’ law depends on the angle between polarization direction and the polarizer’s transmission ______.

The axis sets the direction that passes through. The projection of the electric field onto this axis determines transmitted amplitude.

Explanation

The axis sets the direction that passes through. The projection of the electric field onto this axis determines transmitted amplitude.

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11. Unpolarized light of intensity 100 units passes an ideal polarizer, then a second polarizer at 60° to the first. Final intensity is:

50 units

12.5 units

25 units

6.25 units

First polarizer gives 50 units. Second gives (50 \times \cos^2 60^\circ = 50 \times 1/4 = 12.5).

Explanation

First polarizer gives 50 units. Second gives (50 \times \cos^2 60^\circ = 50 \times 1/4 = 12.5).

12. Malus’ law predicts intensity depends on (\cos^2\theta), so it is always non-negative.

True

False

Squaring removes sign, and intensity can’t be negative physically. This matches the idea that intensity measures energy flow.

Explanation

Squaring removes sign, and intensity can’t be negative physically. This matches the idea that intensity measures energy flow.

13. If you rotate the second polarizer slowly from 0° to 90° relative to the first, the transmitted intensity:

Stays constant

Decreases smoothly to zero

Jumps randomly

Becomes negative

(\cos^2\theta) changes smoothly with angle. It reaches zero at 90° for ideal polarizers.

Explanation

(\cos^2\theta) changes smoothly with angle. It reaches zero at 90° for ideal polarizers.

14. Malus’ law applies to initially polarized light; unpolarized light requires the extra “half” step first.

True

False

Unpolarized light must be treated as containing all polarization directions. The first polarizer sets polarization and halves intensity on average.

Explanation

Unpolarized light must be treated as containing all polarization directions. The first polarizer sets polarization and halves intensity on average.

15. For (\theta = 30^\circ), (\cos^2\theta) is closest to:

0.25

0.50

0.75

1.00

(\cos 30^\circ \approx 0.866). Squaring gives about 0.75.

Explanation

(\cos 30^\circ \approx 0.866). Squaring gives about 0.75.

16. Polarized light of intensity 20 units passes a polarizer at 30°. Output intensity is closest to:

5

10

15

20

With (\cos^2 30^\circ \approx 0.75), (i \approx 20 \times 0.75 = 15). This uses a common-angle approximation.

Explanation

With (\cos^2 30^\circ \approx 0.75), (i \approx 20 \times 0.75 = 15). This uses a common-angle approximation.

17. If you insert a third polarizer between two crossed polarizers, some light can get through.

True

False

The middle polarizer rotates the polarization step-by-step, so the last polarizer isn’t blocking everything. This is a classic counterintuitive polarization result.

Explanation

The middle polarizer rotates the polarization step-by-step, so the last polarizer isn’t blocking everything. This is a classic counterintuitive polarization result.

18. The “cosine-squared” rule comes from projecting the electric field onto the transmission axis and then squaring to get ______.

Polarizers act on field amplitude via projection. Intensity is proportional to amplitude squared, giving the (\cos^2\theta) dependence.

Explanation

Polarizers act on field amplitude via projection. Intensity is proportional to amplitude squared, giving the (\cos^2\theta) dependence.

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19. If the angle doubles from 30° to 60°, the transmitted fraction:

Increases

Decreases

Stays the same

Becomes exactly half always

Increasing angle reduces the transmitted fraction. (\cos^2 (30°)~ 0.75) while (\cos^2 (60°)= 0.25).

Explanation

Increasing angle reduces the transmitted fraction. (\cos^2 (30°)~ 0.75) while (\cos^2 (60°)= 0.25).

20. Polarization calculations can be done without complex maths if you memorise a few key angle values (0°, 45°, 60°, 90°).

True

False

Many problems use common angles with simple (\cos^2) values. This keeps calculations quick and avoids heavy maths.

Explanation

Many problems use common angles with simple (\cos^2) values. This keeps calculations quick and avoids heavy maths.

Malus Law Quiz: Test Your Knowledge Of Polarized Light

1. Malus’ law relates transmitted intensity through a polarizer to:

2.

What first name or nickname would you like us to use?

2. If polarized light is aligned with the polarizer axis, transmission is maximum.

3. If two ideal polarizers are crossed at 90°, the transmitted intensity (for initially polarized light aligned to the first) is:

4. For unpolarized light, the first ideal polarizer transmits about ______ of the intensity.

5. Malus’ law is commonly written as:

6. If (\theta = 60^\circ), then (\cos^2\theta =):

7. Polarized light of intensity 8 units passes a polarizer at 60°. Output intensity is:

8. At (\theta = 45^\circ), Malus’ law predicts half the intensity passes for initially polarized light.

9. If polarized light passes through a polarizer at 0°, then:

10. Malus’ law depends on the angle between polarization direction and the polarizer’s transmission ______.

11. Unpolarized light of intensity 100 units passes an ideal polarizer, then a second polarizer at 60° to the first. Final intensity is:

12. Malus’ law predicts intensity depends on (\cos^2\theta), so it is always non-negative.

13. If you rotate the second polarizer slowly from 0° to 90° relative to the first, the transmitted intensity:

14. Malus’ law applies to initially polarized light; unpolarized light requires the extra “half” step first.

15. For (\theta = 30^\circ), (\cos^2\theta) is closest to:

16. Polarized light of intensity 20 units passes a polarizer at 30°. Output intensity is closest to:

17. If you insert a third polarizer between two crossed polarizers, some light can get through.

18. The “cosine-squared” rule comes from projecting the electric field onto the transmission axis and then squaring to get ______.

19. If the angle doubles from 30° to 60°, the transmitted fraction:

20. Polarization calculations can be done without complex maths if you memorise a few key angle values (0°, 45°, 60°, 90°).