In a system of two polarisers , it is found that the intensity of light from the second polarized is half from that of the first polariser .The angle between their pass axes is

Options

A. `60^@`
B. `30^@`
C. `0^@`
D. `45^@`

Answer

To solve the problem, we will use Malus's Law, which states that when polarized light passes through a polarizer, the intensity of the transmitted light (I) is related to the intensity of the incident light (I0) and the angle (θ) between the light's polarization direction and the axis of the polarizer by the formula:

\[ I = I_0 \cos^2(\theta) \]

### Step-by-Step Solution:

1. **Identify the Given Information:**
- The intensity of light coming out of the second polarizer (I) is half of the intensity coming out of the first polarizer (I0).
- Therefore, we can write:
\[ I = \frac{1}{2} I_0 \]

2. **Apply Malus's Law:**
- According to Malus's Law, the intensity after the second polarizer can be expressed as:
\[ I = I_0 \cos^2(\theta) \]
- Substituting the expression for I from step 1, we have:
\[ \frac{1}{2} I_0 = I_0 \cos^2(\theta) \]

3. **Simplify the Equation:**
- We can cancel \( I_0 \) from both sides (assuming \( I_0 \neq 0 \)):
\[ \frac{1}{2} = \cos^2(\theta) \]

4. **Solve for Cosine:**
- Taking the square root of both sides gives:
\[ \cos(\theta) = \frac{1}{\sqrt{2}} \]

5. **Find the Angle θ:**
- The angle θ can be determined by taking the inverse cosine:
\[ \theta = \cos^{-1}\left(\frac{1}{\sqrt{2}}\right) \]
- This corresponds to:
\[ \theta = 45^\circ \]

6. **Conclusion:**
- The angle between the pass axes of the two polarizers is \( 45^\circ \).

### Final Answer:
The angle between the pass axes of the two polarizers is \( 45^\circ \).