JEE MAIN - Physics (2018 - 16th April Morning Slot - No. 4)
Unpolarized light of intensity I is incident on a system of two polarizers, A followed by B. The intensity of emergent light is I/2. If a third polarizer C is placed between A and B, the intensity of emergent light is reduced to I/3. The angle between the polarizers A and C is $$\theta $$. Then :
cos$$\theta $$ = $${\left( {{2 \over 3}} \right)^{{\raise0.5ex\hbox{$\scriptstyle 1$}
\kern-0.1em/\kern-0.15em
\lower0.25ex\hbox{$\scriptstyle 2$}}}}$$
cos$$\theta $$ = $${\left( {{2 \over 3}} \right)^{{\raise0.5ex\hbox{$\scriptstyle 1$}
\kern-0.1em/\kern-0.15em
\lower0.25ex\hbox{$\scriptstyle 4$}}}}$$
cos$$\theta $$ = $${\left( {{1 \over 3}} \right)^{{\raise0.5ex\hbox{$\scriptstyle 1$}
\kern-0.1em/\kern-0.15em
\lower0.25ex\hbox{$\scriptstyle 2$}}}}$$
cos$$\theta $$ = $${\left( {{1 \over 3}} \right)^{{\raise0.5ex\hbox{$\scriptstyle 1$}
\kern-0.1em/\kern-0.15em
\lower0.25ex\hbox{$\scriptstyle 4$}}}}$$
Explanation
As intensity of emergent beam is reduced to half after passing through two polarisers. It means angle between A and B is 0$$^\circ$$.
_16th_April_Morning_Slot_en_4_1.png)
Now, on placing polariser C between A and B.
_16th_April_Morning_Slot_en_4_2.png)
Intensity after passing through A is $${I_A} = {I \over 2}$$.
Let $$\theta$$ be the angle between A and C. Intensity of light after passing through C is given by
$${I_C} = {I \over 2}{\cos ^2}\theta $$
Intensity of light after passing through polariser B is $${I \over 3}$$.
Angle between C and B is also $$\theta$$ as A is parallel to B.
So, $${I \over 3} = {I_C}{\cos ^2}\theta = {I \over 2}{\cos ^2}\theta \,.\,{\cos ^2}\theta $$
$${\cos ^4}\theta = {2 \over 3} \Rightarrow \cos \theta = {\left( {{2 \over 3}} \right)^{1/4}}$$
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