A) \[\frac{27}{128}\]
B) \[\frac{128}{27}\]
C) \[\frac{37}{128}\]
D) \[\frac{128}{37}\]
Correct Answer: A
Solution :
Let \[{{I}_{0}}\] be the intensity of unpolarized light, then intensity of light transmitted From 1st polarizing sheets\[=\frac{{{I}_{0}}}{2}\] From 2nd polarizing sheet, \[I'=\frac{{{I}_{0}}}{2}(\cos {{30}^{o}})=\frac{{{I}_{0}}}{2}{{\left( \frac{\sqrt{3}}{2} \right)}^{2}}=\frac{{{I}_{0}}}{2}\left( \frac{3}{4} \right)\] From 3rd polarizing sheet, \[I''=I'{{(\cos {{30}^{o}})}^{2}}=\frac{{{I}_{0}}}{2}\left( \frac{3}{4} \right){{\left( \frac{\sqrt{3}}{2} \right)}^{2}}=\frac{{{I}_{0}}}{2}{{\left( \frac{3}{4} \right)}^{2}}\] From 4th polarizing sheet, \[I'''=I''(cos{{30}^{o}})=\frac{{{I}_{0}}}{2}{{\left( \frac{3}{4} \right)}^{2}}={{\left( \frac{\sqrt{3}}{2} \right)}^{2}}={{I}_{0}}\times \frac{27}{128}\] \[\therefore \] \[\frac{I'''}{{{I}_{0}}}=\frac{27}{128}\]
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