Question:
The percentage decrease in the intensity of polarized light when it is passed through an analyser at an angle of \(60^\circ\) is
The percentage decrease in the intensity of polarized light when it is passed through an analyser at an angle of \(60^\circ\) is
Show Hint
For polarized light passing through an analyser,
\[
I=I_0\cos^2\theta.
\]
At
\[
\theta=60^\circ,
\]
\[
I=\frac{I_0}{4}.
\]
So \(25\%\) is transmitted and \(75\%\) is lost.
Updated On: Jul 29, 2026
- \(25\%\)
- \(50\%\)
- \(75\%\)
- \(60\%\)
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The Correct Option is C
Solution and Explanation
Concept:
According to Malus' Law,
\[
I=I_0\cos^2\theta,
\]
where
\[
I_0=\text{incident intensity},
\qquad
I=\text{transmitted intensity},
\]
and \(\theta\) is the angle between the transmission axes of the polarizer and analyser.
Step 1: Apply Malus' Law for \(\theta=60^\circ\). \[ I = I_0\cos^2 60^\circ. \] \[ = I_0\left(\frac12\right)^2. \] \[ = \frac{I_0}{4}. \] Thus, only \[ 25\% \] of the original intensity is transmitted.
Step 2: Calculate the percentage decrease in intensity. Decrease in intensity: \[ I_0-I = I_0-\frac{I_0}{4}. \] \[ = \frac{3I_0}{4}. \] Therefore, \[ \text{Percentage decrease} = \frac{\frac{3I_0}{4}}{I_0}\times100. \] \[ = 75\%. \] Hence, \[ \boxed{\text{Percentage decrease}=75\%} \] \[ \boxed{\text{Answer = (C)}} \]
Step 1: Apply Malus' Law for \(\theta=60^\circ\). \[ I = I_0\cos^2 60^\circ. \] \[ = I_0\left(\frac12\right)^2. \] \[ = \frac{I_0}{4}. \] Thus, only \[ 25\% \] of the original intensity is transmitted.
Step 2: Calculate the percentage decrease in intensity. Decrease in intensity: \[ I_0-I = I_0-\frac{I_0}{4}. \] \[ = \frac{3I_0}{4}. \] Therefore, \[ \text{Percentage decrease} = \frac{\frac{3I_0}{4}}{I_0}\times100. \] \[ = 75\%. \] Hence, \[ \boxed{\text{Percentage decrease}=75\%} \] \[ \boxed{\text{Answer = (C)}} \]
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