A beam of unpolarised light of intensity `I_(0)` is passed through a polaroid "A" and then through another polaroid "B" which is oriented so that its principal plane makes an angle of `45^(@)` relative to that of "A" .The intensity of emergent light is:

Options

A. `I_(0)//2`
B. `I_(0)//8`
C. `I_(0)`
D. `I_(0)//4`

Answer

To solve the problem, we will follow these steps:

### Step 1: Understanding the Initial Conditions
We have a beam of unpolarized light with an intensity \( I_0 \) passing through two polaroids, A and B. The polaroid A will polarize the light, and then polaroid B will further modify the intensity based on its orientation.

**Hint:** Remember that unpolarized light has equal intensity in all directions, and a polaroid only allows light oscillating in its principal plane to pass through.

### Step 2: Applying Malus's Law for Polaroid A
When unpolarized light passes through the first polaroid (A), the intensity of the light that emerges is given by:
\[ I_A = \frac{I_0}{2} \]
This is because a polaroid reduces the intensity of unpolarized light by half.

**Hint:** Malus's Law states that the intensity of polarized light after passing through a polaroid is proportional to the cosine square of the angle between the light's polarization direction and the polaroid's axis.

### Step 3: Applying Malus's Law for Polaroid B
Now, the light emerging from polaroid A is polarized. When this light passes through the second polaroid (B), which is oriented at an angle of \( 45^\circ \) to the first polaroid, we apply Malus's Law again:
\[ I_B = I_A \cdot \cos^2(\theta) \]
where \( \theta = 45^\circ \).

Substituting the value from step 2:
\[ I_B = \frac{I_0}{2} \cdot \cos^2(45^\circ) \]

**Hint:** Remember that \( \cos(45^\circ) = \frac{1}{\sqrt{2}} \).

### Step 4: Calculating \( \cos^2(45^\circ) \)
Calculating \( \cos^2(45^\circ) \):
\[ \cos^2(45^\circ) = \left(\frac{1}{\sqrt{2}}\right)^2 = \frac{1}{2} \]

**Hint:** Squaring the cosine value will give you the fraction of intensity that passes through the second polaroid.

### Step 5: Final Calculation of Intensity
Now substituting back into the equation for \( I_B \):
\[ I_B = \frac{I_0}{2} \cdot \frac{1}{2} = \frac{I_0}{4} \]

**Hint:** This final result shows how much of the original intensity remains after passing through both polaroids.

### Conclusion
The intensity of the emergent light after passing through both polaroids is:
\[ \boxed{\frac{I_0}{4}} \]