Two polaroide `A` and `B` are placed in such a way that the pass -axis of polaroids are perpendicular to each other.Now,another polaroid `C` is placed between `A` and `B` bisecting angle between them.If intensity of unpolarized light is `l_(0)` then intensity of transmitted light atter passing through polaroid `B` will be:

Options

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

Answer

To solve the problem step by step, we will analyze the situation with the three polaroids and apply Malus's Law.

### Step 1: Understanding the Setup
We have three polaroids: A, B, and C. The pass-axis of polaroids A and B are perpendicular to each other (90 degrees apart), and polaroid C is placed between them, bisecting the angle between A and B. This means that the angle between the pass-axis of polaroid A and polaroid C is 45 degrees, and the same applies to the angle between polaroid C and polaroid B.

### Step 2: Intensity after Polaroid A
The initial light is unpolarized with intensity \( I_0 \). When unpolarized light passes through a polaroid, the intensity of the transmitted light is given by:

\[ I_A = \frac{I_0}{2} \]

This is because the intensity of unpolarized light is halved when it passes through the first polaroid.

### Step 3: Intensity after Polaroid C
Now, the light that exits polaroid A is polarized and has intensity \( I_A = \frac{I_0}{2} \). When this light passes through polaroid C, which is at an angle of 45 degrees to the axis of A, we can apply Malus's Law:

\[ I_C = I_A \cos^2(45^\circ) \]

Substituting the values:

\[ I_C = \frac{I_0}{2} \cdot \cos^2(45^\circ) = \frac{I_0}{2} \cdot \left(\frac{1}{\sqrt{2}}\right)^2 = \frac{I_0}{2} \cdot \frac{1}{2} = \frac{I_0}{4} \]

### Step 4: Intensity after Polaroid B
Next, the light that exits polaroid C (with intensity \( I_C = \frac{I_0}{4} \)) now passes through polaroid B, which is also at an angle of 45 degrees to the axis of C. Again, we apply Malus's Law:

\[ I_B = I_C \cos^2(45^\circ) \]

Substituting the values:

\[ I_B = \frac{I_0}{4} \cdot \cos^2(45^\circ) = \frac{I_0}{4} \cdot \left(\frac{1}{\sqrt{2}}\right)^2 = \frac{I_0}{4} \cdot \frac{1}{2} = \frac{I_0}{8} \]

### Final Answer
Thus, the intensity of the transmitted light after passing through polaroid B is:

\[ I_B = \frac{I_0}{8} \]