Question

' $n$ ' polarizing sheets are arranged such that each makes an angle $45^{\circ}$ with the preceeding sheet An unpolarized light of intensity I is incident into this arrangement The output intensity is found to be $I / 64$ The value of $n$ will be:

• A. 6
• B. 5
• C. 3
• D. 4

Answer 1

The Correct Option is A

To solve this problem, we need to apply the concept of polarization. When unpolarized light passes through a series of polarizing sheets, its intensity is reduced according to Malus's law. Malus's law states that when polarized light is incident on a polarizer, the intensity of light transmitted through the polarizer is given by:

I' = I_0 \cos^2 \theta

where I_0 is the initial intensity of the polarized light before hitting the polarizer, \theta is the angle between the light's polarization direction and the polarizer's axis, and I' is the transmitted intensity.

In this case, the light is initially unpolarized, so the light's intensity after passing through the first sheet is reduced by half, because it will only allow half of the incident unpolarized light to pass through. This yields:

I_1 = \frac{I}{2}

Then, each subsequent sheet makes an angle of 45^{\circ} with the preceding one. The intensity after each sheet (I_2, I_3, \ldots, I_n) will be given as:

I_{n} = I_{n-1} \cos^2 45^{\circ}

Since \cos 45^{\circ} = \frac{1}{\sqrt{2}}, it follows that:

I_{n} = I_{n-1} \times \left( \frac{1}{2} \right)

Initially, we have:

I_1 = \frac{I}{2} after the first sheet.

After the second sheet:

I_2 = I_1 \times \left( \frac{1}{2} \right) = \frac{I}{2} \times \frac{1}{2} = \frac{I}{4}

After the third sheet:

I_3 = I_2 \times \left( \frac{1}{2} \right) = \frac{I}{4} \times \frac{1}{2} = \frac{I}{8}

It follows that after the n^{th} sheet:

I_n = \frac{I}{2} \times \left( \frac{1}{2} \right)^{n-1} = \frac{I}{2^n}

We know the output intensity is \frac{I}{64}:

\frac{I}{2^n} = \frac{I}{64}

Simplifying gives:

2^n = 64

Since 64 = 2^6, we have n = 6.

Thus, the correct option is 6.