Plane polarised light is passed through a polaroid. On viewing through the polaroid we find that when the polaroid is given one complete rotation about the direction of light

Options

A. the intensity of light gradually decreases to zero and remains at zero
B. the intensity of light gradually increases to a maximum and remains at maximum
C. there is no change in intensity
D. the intensity of light is twice maximum and twice zero

Answer

To solve the problem of how the intensity of plane polarized light changes as a polaroid is rotated through one complete rotation, we can follow these steps:

### Step-by-Step Solution:

1. **Understanding the Setup**:
- We have plane polarized light passing through a polaroid. The intensity of light passing through a polaroid is given by Malus's Law, which states that the intensity \( I \) of light after passing through a polaroid is given by:
\[ I = I_0 \cos^2 \theta \]
where \( I_0 \) is the initial intensity of the polarized light and \( \theta \) is the angle between the light's polarization direction and the axis of the polaroid.

2. **Analyzing the Rotation**:
- When the polaroid is rotated, \( \theta \) changes from \( 0 \) to \( 2\pi \) (or \( 0^\circ \) to \( 360^\circ \)).
- At \( \theta = 0 \) (or \( 0^\circ \)), the light is aligned with the polaroid, and the intensity is maximum:
\[ I = I_0 \cos^2(0) = I_0 \]
- At \( \theta = \frac{\pi}{2} \) (or \( 90^\circ \)), the light is perpendicular to the polaroid's axis, and the intensity is zero:
\[ I = I_0 \cos^2\left(\frac{\pi}{2}\right) = 0 \]
- At \( \theta = \pi \) (or \( 180^\circ \)), the light is again aligned with the polaroid, and the intensity returns to maximum:
\[ I = I_0 \cos^2(\pi) = I_0 \]
- At \( \theta = \frac{3\pi}{2} \) (or \( 270^\circ \)), the intensity is again zero:
\[ I = I_0 \cos^2\left(\frac{3\pi}{2}\right) = 0 \]
- Finally, at \( \theta = 2\pi \) (or \( 360^\circ \)), the intensity is maximum again:
\[ I = I_0 \cos^2(2\pi) = I_0 \]

3. **Identifying the Pattern**:
- As we rotate the polaroid from \( 0 \) to \( 2\pi \):
- The intensity goes from maximum (\( I_0 \)) to zero (\( 0 \)) at \( \frac{\pi}{2} \),
- Back to maximum (\( I_0 \)) at \( \pi \),
- Down to zero (\( 0 \)) at \( \frac{3\pi}{2} \),
- And finally back to maximum (\( I_0 \)) at \( 2\pi \).
- Therefore, in one complete rotation, the intensity goes from maximum to zero and back to maximum, twice.

4. **Conclusion**:
- The correct answer is that the intensity of the light varies such that it reaches maximum intensity twice and zero intensity twice during one complete rotation of the polaroid.

### Final Answer:
The intensity of the light is twice maximum and twice zero.