A point source \(S\) emits unpolarized light uniformly in all directions. At two points A and B, the ratio \(r = I_A/I_B\) of the intensities of light is \(2\). If a set of two polaroids having 45° angle between their pass-axes is placed just before point B, then the new value of r will be ____
Correct Answer: 8
Approach Solution - 1
Case-I: Without Polaroids
The source emits unpolarised light, and the given intensity ratio is:
\[ r = \frac{I_A}{I_B} = 2 \]
Case-II: With Polaroids at Point B
When the first polaroid is introduced, the intensity of light becomes:
\[ I'_B = \frac{I_B}{2} \]
When the second polaroid is introduced at 45° relative to the first, the intensity of light becomes:
\[ I''_B = I'_B \cdot \cos^2(45^\circ) = \frac{I_B}{2} \cdot \frac{1}{2} = \frac{I_B}{4} \]
New Intensity Ratio
The new intensity ratio is:
\[ r' = \frac{I_A}{I''_B} = \frac{I_A}{\frac{I_B}{4}} = 4 \cdot \frac{I_A}{I_B} = 4 \times 2 = 8 \]
Final Answer:
r = 8
Approach Solution -2
To solve the problem, analyze the intensity changes due to distance and polaroids.
Given:
- Point source \(S\) emits unpolarized light uniformly.
- Intensity ratio at points \(A\) and \(B\):
\[
r = \frac{I_A}{I_B} = 2
\]
- Two polaroids with 45° angle between their pass-axes are placed before point \(B\).
Step 1: Intensity ratio without polaroids:
Since source is uniform and unpolarized, intensity decreases with distance squared:
\[
I \propto \frac{1}{d^2}
\]
Thus,
\[
\frac{I_A}{I_B} = \frac{d_B^2}{d_A^2} = 2
\]
or equivalently,
\[
\frac{I_B}{I_A} = \frac{1}{2}
\]
Step 2: Effect of two polaroids on intensity:
- First polaroid reduces intensity of unpolarized light by half:
\[
I' = \frac{I_B}{2}
\]
- Second polaroid at angle \(\theta = 45^\circ\) passes intensity according to Malus's law:
\[
I'' = I' \cos^2 45^\circ = \frac{I_B}{2} \times \left(\frac{1}{\sqrt{2}}\right)^2 = \frac{I_B}{2} \times \frac{1}{2} = \frac{I_B}{4}
\]
Step 3: New intensity at point B after polaroids:
\[
I_B' = \frac{I_B}{4}
\]
Step 4: New ratio \(r'\) of intensities at \(A\) and \(B\):
\[
r' = \frac{I_A}{I_B'} = \frac{I_A}{I_B / 4} = 4 \times \frac{I_A}{I_B} = 4 \times 2 = 8
\]
Final Answer:
\[
\boxed{8}
\]
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