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