JEE MAIN - Physics (2014 (Offline) - No. 10)
Two beams, $$A$$ and $$B$$, of plane polarized light with mutually perpendicular planes of polarization are seen through a polaroid. From the position when the beam $$A$$ has maximum intensity (and beam $$B$$ has zero intensity), a rotation of polaroid through $${30^ \circ }$$ makes the two beams appear equally bright. If the initial intensities of the two beams are $${{\rm I}_A}$$ and $${{\rm I}_B}$$ respectively, then $${{{{\rm I}_A}} \over {{{\rm I}_B}}}$$ equals:
$$3$$
$${3 \over 2}$$
$$1$$
$${1 \over 3}$$
Explanation
According to malus law, intensity of emerging beam is given by,
$$I = {I_0}{\cos ^2}\theta $$
Now, $${I_{A'}} = {I_A}{\cos ^2}{30^ \circ }$$
$${I_{B'}} = {I_B}{\cos ^2}{60^ \circ }$$
As $${I_{A'}} = {I_{B'}}$$
$$ \Rightarrow {I_A} \times {3 \over 4} = {I_B} \times {1 \over 4}$$
$$\therefore$$ $${{{I_A}} \over {{I_B}}} = {1 \over 3}$$
$$I = {I_0}{\cos ^2}\theta $$
Now, $${I_{A'}} = {I_A}{\cos ^2}{30^ \circ }$$
$${I_{B'}} = {I_B}{\cos ^2}{60^ \circ }$$
As $${I_{A'}} = {I_{B'}}$$
$$ \Rightarrow {I_A} \times {3 \over 4} = {I_B} \times {1 \over 4}$$
$$\therefore$$ $${{{I_A}} \over {{I_B}}} = {1 \over 3}$$
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