Two polaroids are kept crossed to each other. Now one of them is rotated through an angle of `45^(@)` . The percentage of incident light now transmitted through the system is

Options

A. 0.15
B. 0.25
C. 0.5
D. 0.6

Explanation

To solve the problem, we will follow these steps:

### Step 1: Understand the setup
We have two polaroids, initially crossed (i.e., oriented at 90 degrees to each other). When light passes through the first polaroid, its intensity is reduced.

### Step 2: Calculate the intensity after the first polaroid
Let the intensity of the incident light be \( I_0 \). When light passes through the first polaroid, the intensity is reduced to half:
\[ I_1 = \frac{I_0}{2} \]

### Step 3: Apply Malus's Law for the second polaroid
Now, the second polaroid is rotated by \( 45^\circ \). According to Malus's Law, the intensity of light transmitted through a polaroid is given by:
\[ I = I_1 \cdot \cos^2(\theta) \]
where \( \theta \) is the angle between the light's polarization direction and the axis of the polaroid.

In this case, after rotating the second polaroid by \( 45^\circ \), we have:
\[ I_2 = I_1 \cdot \cos^2(45^\circ) \]

### Step 4: Substitute the values
We already found \( I_1 = \frac{I_0}{2} \) and \( \cos(45^\circ) = \frac{1}{\sqrt{2}} \):
\[ I_2 = \frac{I_0}{2} \cdot \left(\frac{1}{\sqrt{2}}\right)^2 \]
\[ I_2 = \frac{I_0}{2} \cdot \frac{1}{2} = \frac{I_0}{4} \]

### Step 5: Calculate the percentage of transmitted light
To find the percentage of incident light transmitted through the system, we calculate:
\[ \frac{I_2}{I_0} = \frac{I_0/4}{I_0} = \frac{1}{4} \]
This means that \( 25\% \) of the incident light is transmitted through the system.

### Final Answer
The percentage of incident light transmitted through the system is \( 25\% \).

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