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If 10% of intensity is passed from analy...

If 10% of intensity is passed from analyser, then, the angle by which analyser should be rotated such that transmitted intensity becomes zero. (Assume no absorption by analyser and polarizer).

A

`60^(@)`

B

`18.4^(@)`

C

`45^(@)`

D

`71.6^(@)`

Text Solution

AI Generated Solution

The correct Answer is:
To solve the problem, we will follow these steps: ### Step 1: Understand the Problem We need to find the angle by which the analyzer should be rotated so that the transmitted intensity becomes zero. We know that 10% of the intensity is transmitted through the analyzer initially. ### Step 2: Use Malus's Law According to Malus's Law, the transmitted intensity \( I \) through a polarizer and analyzer is given by: \[ I = I_0 \cos^2 \theta \] where \( I_0 \) is the initial intensity and \( \theta \) is the angle between the light's polarization direction and the axis of the analyzer. ### Step 3: Set Up the Equation Given that 10% of the intensity is transmitted, we can express this as: \[ I = \frac{I_0}{10} \] Substituting this into Malus's Law gives: \[ \frac{I_0}{10} = I_0 \cos^2 \theta \] ### Step 4: Simplify the Equation We can cancel \( I_0 \) from both sides (assuming \( I_0 \neq 0 \)): \[ \frac{1}{10} = \cos^2 \theta \] ### Step 5: Solve for \( \theta \) Taking the square root of both sides, we find: \[ \cos \theta = \frac{1}{\sqrt{10}} \] Now, we can calculate \( \theta \): \[ \theta = \cos^{-1}\left(\frac{1}{\sqrt{10}}\right) \approx 71.6^\circ \] ### Step 6: Determine the Angle for Zero Intensity To find the angle at which the transmitted intensity becomes zero, we set: \[ I = I_0 \cos^2 \theta = 0 \] This implies: \[ \cos^2 \theta = 0 \implies \cos \theta = 0 \] Thus: \[ \theta = 90^\circ \] ### Step 7: Calculate the Required Rotation Angle The angle by which the analyzer should be rotated is the difference between the final angle (90°) and the initial angle (71.6°): \[ \Delta \theta = 90^\circ - 71.6^\circ = 18.4^\circ \] ### Final Answer The angle by which the analyzer should be rotated so that the transmitted intensity becomes zero is: \[ \Delta \theta = 18.4^\circ \] ---
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