KCET · Physics · Atomic Physics
A polarized light of intensity \( I_{0} \) is passed through another polarizer whose pass axis makes an
angle of \( 60^{\circ} \) with the pass axis of the former. What is the intensity of emergent polarized light
from second polarizer?
- A \( \mathrm{O} I=I_{0} \)
- B \( I=I_{0} / 6 \)
- C \( O=I_{0} / 5 \)
- D \( I_{0} / 4 \)
Answer & Solution
Correct Answer
(D) \( I_{0} / 4 \)
Step-by-step Solution
Detailed explanation
According to Malus Law, we have
\[
I=I_{0} \cos ^{2} \theta
\]
where, \( I \) is intensity of emergent polarised light; \( I_{0} \) is intensity of incident polarised light; \( \theta \) is angle between pass axis of
both polarisers
Here \( \theta=60^{\circ} \)
Therefore \( I=I_{0}(\cos 60)^{2}=I_{0}\left(\frac{1}{2}\right)^{2}=\frac{I_{0}}{4} \)
Thus, intensity of emergent polarised light from second polarizer \( =\frac{I_{0}}{4} \)
\[
I=I_{0} \cos ^{2} \theta
\]
where, \( I \) is intensity of emergent polarised light; \( I_{0} \) is intensity of incident polarised light; \( \theta \) is angle between pass axis of
both polarisers
Here \( \theta=60^{\circ} \)
Therefore \( I=I_{0}(\cos 60)^{2}=I_{0}\left(\frac{1}{2}\right)^{2}=\frac{I_{0}}{4} \)
Thus, intensity of emergent polarised light from second polarizer \( =\frac{I_{0}}{4} \)
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