Unpolarised light falls on two polarizing sheets placed one on top of the other. What must be the angle between the characteristic directions of the sheets if the intensity of the final transmitted light is one-third the maximum intensity of the first transmitted beam?

Options

A. `cos^(-1)((sqrt2)/(sqrt3))`
B. `cos^(-1)((sqrt3)/(2))`
C. `cos^(-1)((1)/(sqrt3))`
D. `cos^(-1)((1)/(sqrt2))`

Answer

To solve the problem, we need to analyze the situation involving two polarizing sheets and the intensity of light transmitted through them. Here’s a step-by-step solution:

### Step 1: Understand the Initial Conditions
- We have unpolarized light with an initial intensity \( I_0 \).
- The first polarizing sheet will reduce the intensity of the light.

**Hint:** Remember that unpolarized light passing through a polarizer gets its intensity halved.

### Step 2: Calculate the Intensity After the First Polarizer
- When unpolarized light passes through the first polarizer (P1), the intensity of the transmitted light \( I_1 \) is given by:
\[ I_1 = \frac{I_0}{2} \]

**Hint:** The intensity after the first polarizer is always half of the initial intensity for unpolarized light.

### Step 3: Set Up the Condition for the Second Polarizer
- The problem states that the final transmitted intensity \( I_2 \) is one-third of the maximum intensity of the first transmitted beam \( I_1 \). Thus:
\[ I_2 = \frac{1}{3} I_1 = \frac{1}{3} \left( \frac{I_0}{2} \right) = \frac{I_0}{6} \]

**Hint:** Use the relationship given in the problem to express \( I_2 \) in terms of \( I_0 \).

### Step 4: Apply Malus's Law for the Second Polarizer
- The intensity after passing through the second polarizer (P2) is given by Malus's Law:
\[ I_2 = I_1 \cos^2 \theta \]
where \( \theta \) is the angle between the transmission axes of the two polarizers.

- Substituting the expression for \( I_1 \):
\[ \frac{I_0}{6} = \left( \frac{I_0}{2} \right) \cos^2 \theta \]

**Hint:** Remember that Malus's Law relates the transmitted intensity to the angle between the light and the polarizer's axis.

### Step 5: Simplify the Equation
- Cancelling \( I_0 \) from both sides (assuming \( I_0 \neq 0 \)):
\[ \frac{1}{6} = \frac{1}{2} \cos^2 \theta \]

- Multiplying both sides by 2:
\[ \frac{1}{3} = \cos^2 \theta \]

**Hint:** Isolate \( \cos^2 \theta \) to find the cosine value.

### Step 6: Solve for \( \theta \)
- Taking the square root of both sides:
\[ \cos \theta = \frac{1}{\sqrt{3}} \]

- Therefore, the angle \( \theta \) is:
\[ \theta = \cos^{-1} \left( \frac{1}{\sqrt{3}} \right) \]

**Hint:** Use the inverse cosine function to find the angle.

### Final Answer
The angle between the characteristic directions of the sheets is:
\[ \theta = \cos^{-1} \left( \frac{1}{\sqrt{3}} \right) \]