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GUJCET · Physics · Wave Optics
A plane polarized light is incident normally on a tourmaline plate. Its \(\vec{E}\) vectors make an angle of \(60^{\circ}\) with the optic axis of the plate. Find the percentage difference between initial and final intensities.
- A \(50 \%\)
- B \(25 \%\)
- C \(75 \%\)
- D \(90 \%\)
Answer & Solution
Correct Answer
(C) \(75 \%\)
Step-by-step Solution
Detailed explanation
(C) \(75 \%\)
Let the initial intensity of the light be \(\mathrm{I}_{0}\).
Final intensity of light
\(
\begin{aligned}
& \mathrm{I}=\mathrm{I}_{0} \cos ^{2} \theta \\
\therefore \quad & \mathrm{I}=\mathrm{I}_{0} \cos ^{2} 60^{\circ} \\
\therefore \quad & \mathrm{I}=\frac{\mathrm{I}_{0}}{4}
\end{aligned}
\)
Percentage change in the initial and final intensities
\(
\begin{aligned}
& =\frac{I_{0}-I}{I_{0}} \times 100 \\
& =\frac{I_{0}-\frac{I_{0}}{4}}{I_{0}} \times 100 \\
& =\frac{3 I_{0}}{4 I_{0}} \times 100 \\
& =75 \%
\end{aligned}
\)
Let the initial intensity of the light be \(\mathrm{I}_{0}\).
Final intensity of light
\(
\begin{aligned}
& \mathrm{I}=\mathrm{I}_{0} \cos ^{2} \theta \\
\therefore \quad & \mathrm{I}=\mathrm{I}_{0} \cos ^{2} 60^{\circ} \\
\therefore \quad & \mathrm{I}=\frac{\mathrm{I}_{0}}{4}
\end{aligned}
\)
Percentage change in the initial and final intensities
\(
\begin{aligned}
& =\frac{I_{0}-I}{I_{0}} \times 100 \\
& =\frac{I_{0}-\frac{I_{0}}{4}}{I_{0}} \times 100 \\
& =\frac{3 I_{0}}{4 I_{0}} \times 100 \\
& =75 \%
\end{aligned}
\)
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