Optics 6 Question 1
1. A system of three polarisers $P _1, P _2, P _3$ is set up such that the pass axis of $P _3$ is crossed with respect to that of $P _1$. The pass axis of $P _2$ is inclined at $60^{\circ}$ to the pass axis of $P _3$. When a beam of unpolarised light of intensity $I _0$ is incident on $P _1$, the intensity of light transmitted by the three polarisers is $I$. The ratio $\left(I _0 / I\right)$ equals (nearly)
(2019 Main, 12 April II)
(a) 5.33
(b) 16.00
(c) 10.67
(d) 1.80
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Solution:
When unpolarised light pass through polaroid $P _1$, intensity obtained is
$$ I _1=\frac{I _0}{2} $$
where, $I _0=$ intensity of incident light.
Now, this transmitted light is polarised and it pass through polariser $P _2$. So, intensity $I _2$ transmitted is obtained by Malus law.
$$ \Rightarrow \quad I _2=I _1 \cos ^{2} \theta $$
As angle of pass axis of $P _1$ and $P _3$ is $90^{\circ}$ and angle of pass axis of $P _2$ and $P _3$ is $60^{\circ}$, so angle between pass axis of $P _1$ and $P _2$ is $\left(90^{\circ}-60^{\circ}\right)=30^{\circ}$. So,
When this light pass through third polariser $P _3$, intensity $I$ transmitted is again obtained by Malus law. So, $I=I _2 \cos ^{2} 60^{\circ}=\frac{3}{8} I _0 \quad \cos ^{2} 60^{\circ}$
$$ =\frac{3}{8} I _0 \times \frac{1}{2}^{2}=\frac{3}{32} I _0 $$
So, ratio $\frac{I _0}{I}=\frac{32}{3}=10.67$
2 As we know,
Path difference introduced by thin film,
$$ \Delta=(\mu-1) t $$
and if fringe pattern shifts by one frings width, then path difference,
$$ \Delta=1 \times \lambda=\lambda $$
So, from Eqs. (i) and (ii), we get
$$ (\mu-1) t=\lambda \Rightarrow t=\frac{\lambda}{\mu-1} $$
Alternate Solution
Path difference introduced by the thin film of thickness $t$ and refractive index $\mu$ is given by
$$ \Delta=(\mu-1) t $$
$\therefore$ Position of the fringe is
$$ x=\frac{\Delta D}{d}=\frac{(\mu-1) t D}{d} $$
Fringe width of one fringe is given by
$$ \beta=\frac{\lambda D}{d} $$
Given that $x=\beta$, so from Eqs. (i) and (ii), we get
$$ \begin{array}{ll} \Rightarrow & \frac{(\mu-1) t D}{d}=\frac{\lambda D}{d} \\ \Rightarrow & (\mu-1) t=\lambda \text { or } t=\frac{\lambda}{(\mu-1)} \end{array} $$
