Question 157: Waves Optics

Question

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 $$	heta $$. Then :

Accepted Answer

As intensity of emergent beam is reduced to half after passing through two polarisers. It means angle  between A and B is 0$$^\circ$$.
Now, on placing polariser C between A and B.
Intensity after passing through A is $${I_A} = {I \over 2}$$.
Let $$	heta$$ be the angle between A and C. Intensity of light after passing through C is given by
$${I_C} = {I \over 2}{\cos ^2}	heta $$
Intensity of light after passing through polariser B is $${I \over 3}$$.
Angle between C and B is also $$	heta$$ as A is parallel to B.
So, $${I \over 3} = {I_C}{\cos ^2}	heta  = {I \over 2}{\cos ^2}	heta \,.\,{\cos ^2}	heta $$
$${\cos ^4}	heta  = {2 \over 3} \Rightarrow \cos 	heta  = {\left( {{2 \over 3}} ight)^{1/4}}$$

Answer

cos$$	heta $$ = $${\left( {{2 \over 3}} ight)^{{aise0.5ex\hbox{$\scriptstyle 1$}
\kern-0.1em/\kern-0.15em
\lower0.25ex\hbox{$\scriptstyle 2$}}}}$$

Answer

cos$$	heta $$ = $${\left( {{2 \over 3}} ight)^{{aise0.5ex\hbox{$\scriptstyle 1$}
\kern-0.1em/\kern-0.15em
\lower0.25ex\hbox{$\scriptstyle 4$}}}}$$

Answer

cos$$	heta $$ = $${\left( {{1 \over 3}} ight)^{{aise0.5ex\hbox{$\scriptstyle 1$}
\kern-0.1em/\kern-0.15em
\lower0.25ex\hbox{$\scriptstyle 2$}}}}$$

Answer

cos$$	heta $$ = $${\left( {{1 \over 3}} ight)^{{aise0.5ex\hbox{$\scriptstyle 1$}
\kern-0.1em/\kern-0.15em
\lower0.25ex\hbox{$\scriptstyle 4$}}}}$$

Waves Optics

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 :

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