`n'` polarizing sheets are arranged such that each makes an angle `45^(@)` with the preceeding sheet.An unpolarized light of intensity I is incident into this arrangement.The output intensity is found to be `(1)/(64)`. The value of `n` will be:

Options

A. `6`
B. `5`
C. `3`
D. `4`

Answer

To solve the problem, we need to analyze the effect of multiple polarizing sheets on the intensity of light. Here's a step-by-step solution:

### Step 1: Understand the effect of a single polarizing sheet
When unpolarized light passes through a polarizing sheet, the intensity of the transmitted light is given by:

\[ I' = I \cdot \cos^2(\theta) \]

where \(I\) is the intensity of the incident light, \(I'\) is the intensity after passing through the polarizer, and \(\theta\) is the angle between the light's polarization direction and the axis of the polarizer.

### Step 2: Apply the formula for the first polarizing sheet
In this case, the first polarizing sheet makes an angle of \(45^\circ\) with the incident unpolarized light. Therefore, we have:

\[ I_1 = I \cdot \cos^2(45^\circ) = I \cdot \left(\frac{1}{\sqrt{2}}\right)^2 = I \cdot \frac{1}{2} \]

### Step 3: Generalize for \(n\) polarizing sheets
Each subsequent polarizing sheet also makes an angle of \(45^\circ\) with the preceding sheet. Thus, the intensity after passing through \(n\) sheets can be expressed as:

\[ I_n = I_1 \cdot \cos^2(45^\circ) \cdot \cos^2(45^\circ) \cdots \text{(n times)} \]

This simplifies to:

\[ I_n = I \cdot \left(\frac{1}{2}\right)^n \]

### Step 4: Set up the equation with the given output intensity
We are given that the output intensity \(I_n\) is \(\frac{I}{64}\). Therefore, we can set up the equation:

\[ I \cdot \left(\frac{1}{2}\right)^n = \frac{I}{64} \]

### Step 5: Simplify the equation
Dividing both sides by \(I\) (assuming \(I \neq 0\)) gives us:

\[ \left(\frac{1}{2}\right)^n = \frac{1}{64} \]

### Step 6: Express \(64\) as a power of \(2\)
We know that:

\[ 64 = 2^6 \]

Thus, we can rewrite the equation as:

\[ \left(\frac{1}{2}\right)^n = \frac{1}{2^6} \]

### Step 7: Equate the exponents
Since the bases are the same, we equate the exponents:

\[ -n = -6 \implies n = 6 \]

### Conclusion
The value of \(n\) is \(6\).