Unpolarised light of intensity `I_0` passes through five successive polaroid sheets,each of whose transmission axes makes an angle `45^@`. With the previous one.The intensity of the finally transmitted light is

Options

A. `(I_@)/64`
B. `(I_@)/16`
C. `(I_@)/8`
D. (`I_@)/32`

Answer

To find the intensity of the light transmitted through five successive polaroid sheets, we can follow these steps:

### Step 1: Initial Intensity
The initial intensity of the unpolarized light is given as \( I_0 \).

### Step 2: First Polaroid
When unpolarized light passes through the first polaroid, the intensity of the transmitted light is reduced to half:
\[ I_1 = \frac{I_0}{2} \]

### Step 3: Subsequent Polaroids
Each subsequent polaroid is oriented at an angle of \( 45^\circ \) to the previous one. According to Malus's Law, the intensity of light transmitted through a polaroid is given by:
\[ I_n = I_{n-1} \cos^2(\theta) \]
where \( \theta \) is the angle between the light's polarization direction and the polaroid's axis.

### Step 4: Calculate Intensity After Each Polaroid
1. **Second Polaroid**:
\[ I_2 = I_1 \cos^2(45^\circ) = \frac{I_0}{2} \cdot \left(\frac{1}{\sqrt{2}}\right)^2 = \frac{I_0}{2} \cdot \frac{1}{2} = \frac{I_0}{4} \]

2. **Third Polaroid**:
\[ I_3 = I_2 \cos^2(45^\circ) = \frac{I_0}{4} \cdot \frac{1}{2} = \frac{I_0}{8} \]

3. **Fourth Polaroid**:
\[ I_4 = I_3 \cos^2(45^\circ) = \frac{I_0}{8} \cdot \frac{1}{2} = \frac{I_0}{16} \]

4. **Fifth Polaroid**:
\[ I_5 = I_4 \cos^2(45^\circ) = \frac{I_0}{16} \cdot \frac{1}{2} = \frac{I_0}{32} \]

### Final Result
The intensity of the light transmitted after passing through all five polaroids is:
\[ I_5 = \frac{I_0}{32} \]

### Conclusion
Thus, the final transmitted intensity is \( \frac{I_0}{32} \).

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