When a polaroid sheet is rotated between two crossed polaroids, the intensity of the transmitted will be maximum, when angle `theta` between pass axes is

Options

A. `theta=90^@`
B. `theta=0^@`
C. `theta=45^@`
D. it is independent of angle

Answer

To solve the problem of finding the angle \( \theta \) at which the intensity of transmitted light is maximum when a polaroid sheet is rotated between two crossed polaroids, we can follow these steps:

### Step-by-Step Solution:

1. **Understanding the Setup**:
We have two crossed polaroids (let's call them \( P_1 \) and \( P_2 \)) that are oriented at an angle of \( 90^\circ \) to each other. We introduce a third polaroid \( P_0 \) that is rotated at an angle \( \theta \) with respect to \( P_1 \).

2. **Applying Malus's Law**:
According to Malus's law, the intensity of light transmitted through a polaroid is given by:
\[ I = I_0 \cos^2(\phi) \]
where \( I_0 \) is the intensity of the incoming light and \( \phi \) is the angle between the light's polarization direction and the polaroid's transmission axis.

3. **Calculating Intensity After First Polaroid**:
The intensity of light after passing through \( P_1 \) is:
\[ I_1 = I_0 \cos^2(0) = I_0 \]
(since the light is initially polarized along the axis of \( P_1 \)).

4. **Intensity After the Third Polaroid**:
The intensity after passing through the third polaroid \( P_0 \) is:
\[ I_2 = I_1 \cos^2(\theta) = I_0 \cos^2(\theta) \]

5. **Intensity After the Second Polaroid**:
The angle between \( P_0 \) and \( P_2 \) is \( 90^\circ - \theta \). Thus, the intensity after passing through \( P_2 \) is:
\[ I_3 = I_2 \cos^2(90^\circ - \theta) = I_0 \cos^2(\theta) \sin^2(\theta) \]

6. **Simplifying the Expression**:
We can express \( I_3 \) as:
\[ I_3 = I_0 \cos^2(\theta) \sin^2(\theta) \]
Using the identity \( \sin(2\theta) = 2\sin(\theta)\cos(\theta) \), we can rewrite this as:
\[ I_3 = \frac{I_0}{4} \sin^2(2\theta) \]

7. **Finding Maximum Intensity**:
The maximum value of \( \sin^2(2\theta) \) is 1, which occurs when \( 2\theta = 90^\circ \) or \( \theta = 45^\circ \).

8. **Conclusion**:
Therefore, the angle \( \theta \) at which the intensity of the transmitted light is maximum is:
\[ \theta = 45^\circ \]

### Final Answer:
The angle \( \theta \) between the pass axes for maximum transmitted intensity is \( 45^\circ \).

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