Question

An unpolarized beam of light is incident on a group of four polarizing sheets, which are arranged in such a way that the characteristic direction of each polarizing sheet makes an angle of $30{}^\circ$ with that of the preceding sheet. The fraction of incident unpolarized light transmitted is
(a) $\dfrac{27}{128}$
(b) $\dfrac{128}{27}$
(c) $\dfrac{37}{128}$
(d) $\dfrac{128}{37}$

Answer 1

Hint: First, we should know the formula for unpolarized light transmitted which is given by $\dfrac{I}{{{I}_{0}}}$ . So, to calculate I’ will be a beam passing from one polarizing sheet then multiplying that value with the beam passing through the second sheet. Again, the value of this will be multiplying to the beam passing through the third sheet. Similarly, with the beam passing through the fourth sheet. Final answer obtained is to be kept in the above equation, thus the required answer will get.

Formula used: $I={{I}_{0}}{{\cos }^{2}}\theta$ , $\dfrac{I}{{{I}_{0}}}$

Complete step-by-step answer:
Now, here we are given that sheet makes an angle of $\theta =30{}^\circ$ . Considering variable ${{I}_{0}}$ as intensity of unpolarized light.
So, now finding value of beam passing through first sheet we get as,
The first polarizer reduced intensity of light by factor $\dfrac{1}{2}$ , therefore the intensity of the beam after it passes through first sheet is
${{I}_{1}}=\dfrac{{{I}_{0}}}{2}$ …………………………….(1)
Now, considering bean passing through second sheet we get,
Beam passing through second sheet ${{I}_{2}}={{I}_{1}}\text{ }\times {{\cos }^{2}}30$
Substituting value of equation (1), we get
${{I}_{2}}=\dfrac{{{I}_{0}}}{2}\text{ }{{\cos }^{2}}30$ ……………………(2)
Now, considering bean passing through third sheet we get,
Beam passing through third sheet ${{I}_{3}}={{I}_{2}}\text{ }\times {{\cos }^{2}}30$
Substituting value of equation (2), we get
${{I}_{3}}=\dfrac{{{I}_{0}}}{2}\text{ }{{\cos }^{2}}30\times {{\cos }^{2}}30=\dfrac{{{I}_{0}}}{2}\text{ }{{\cos }^{4}}30$ ………………………(3)
Now, considering bean passing through fourth sheet we get,
Beam passing through third sheet ${{I}_{4}}={{I}_{3}}\text{ }\times {{\cos }^{2}}30$
Substituting value of equation (3), we get
${{I}_{4}}=\dfrac{{{I}_{0}}}{2}\text{ }{{\cos }^{4}}30\times {{\cos }^{2}}30=\dfrac{{{I}_{0}}}{2}\text{ }{{\cos }^{6}}30$ ……………………….(4)
Now using the formula $\dfrac{I}{{{I}_{0}}}$ we get,
$\dfrac{{{I}_{4}}}{{{I}_{0}}}\Rightarrow \dfrac{{{I}_{0}}{{\cos }^{6}}\theta }{2{{I}_{0}}}$
$\dfrac{{{I}_{4}}}{{{I}_{0}}}\Rightarrow \dfrac{{{I}_{0}}{{\cos }^{6}}\theta }{2{{I}_{0}}}=\dfrac{{{\cos }^{6}}30}{2}=\dfrac{{{\left( \dfrac{\sqrt{3}}{2} \right)}^{6}}}{2}$
$=\dfrac{27}{2\left( 64 \right)}=\dfrac{27}{128}$
Thus, the fraction of incident unpolarized light transmitted is $\dfrac{27}{128}$.
Option (a) is correct.

Note: Be careful by using the formula $I={{I}_{0}}{{\cos }^{2}}\theta$ as it involves square of cosine function. Instead of using cosine function, sometimes mistakes happen by using sine function and then on solving we will get incorrect answers. So, don’t make these silly mistakes.