Derivation of interference of light due to thin-film:
Let's consider a Ray of light $AB$ incident on a thin film of thickness $t$ and the refractive index of a thin film is $\mu$
The ray $AB$ is partially reflected and partially transmitted at $B$. The transmitted BC is against partially transmitted and partially reflected at $C$. The reflected ray $CD$ is partially reflected and partially refracted at $D$.
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| Propagation of light ray in thin film |
The interference pattern in reflected light will be due to ray $BF$ and $DH$ which are coherent as they are both derived from the same Ray $AB$.
The interference pattern in transmitted light will be due to ray $CI$ and $EJ$.
The path difference between $BF$ and $DH$ ray will be
$\Delta=\mu(BC+CD)-BG \qquad(1)$
The triangle $\Delta BCK$ and $\Delta CDK$ are congruent because
$\begin{Bmatrix}
BK=KD
\\
BC=CD
\\
CK=t \end{Bmatrix} \qquad (2)$
From equation $(1)$ and $(2)$
$\Delta=2 \mu BC-BG \qquad(3)$
In $\Delta BCK$
$cos \: r= \frac{CK}{BC}$
$BC= \frac{t}{cos \: r} \qquad(4)$
In $\Delta BGD$
$sin \: i= \frac{BG}{BD}$
$BG= BD \: sin\: i $
$BG= (BK+KD)\: sin\: i $
$BG= 2BK \: sin\: i \qquad(5)$ {From eqaution $(2)$}
Again In $\Delta BCK$
$tan \:r = \frac{BK}{CK}$
$BK=t.tan\:r \qquad(6)$
Put the value of $BK$ in equation $(5)$
$BG=2t\:sin \:i .tan\: r$
$BG=2t\:sin \:i .\frac{sin \: r}{cos \: r}$
$BG= 2t \frac{sin \:i}{sin \: r} \frac{sin^{2} \:r}{cos \: r}$
$BG=2\mu t \frac{sin^{2} \:r}{cos \: r} \qquad (7)\qquad \left( \mu =\frac{sin \:i}{sin \: r}\right)$
Now put the value of $BG$ and $BC$ in equation $(3)$
$\Delta= \frac{2 \mu t}{cos \: r}-2\mu t\frac{sin^{2} \:r}{cos \: r}$
$\Delta=\frac{2 \mu t}{cos \:r} \left[1-sin^{2} \:r \right]$
$\Delta=2 \mu t \frac{cos^{2} \:r}{cos \:r} $
$\Delta=2 \mu t \: cos \:r$
Interference in a reflected ray:
The ray $BF$ suffers phase change of $\pi$ due to reflection from the denser medium at $B$. Therefore the path difference of $\frac{\lambda}{2}$ is introduced between two rays due to reflection-
$\Delta= 2\mu t \: cos\:r \pm \frac{\lambda}{2}$
Constructive Interference due reflected ray:
for constructive interference
$\Delta=n \lambda$
so from the above equations, we get
$2\mu t \: cos\: r \pm \frac{\lambda}{2}=n \lambda$
$2\mu t \: cos\: r = (2n \pm 1)\frac{\lambda}{2}$
Destructive Interference due to reflected ray:
for destructive interference
$\Delta=(2n \pm 1)\frac{\lambda}{2}$
so from the above equations, we get
$2\mu t \: cos\: r \pm \frac{\lambda}{2}=(2n \pm 1)\frac{\lambda}{2}$
$2\mu t \: cos\: r =n \lambda$
Interference in transmitted ray:
The ray $CI$ and $EJ$ in transmitted Ray have the same path difference as a reflected ray. There is not any change in phase for $CI$ due to reflection as it gets transmitted at $C$. The ray $EJ$ also does not undergo phase change due to reflection as it is reflected from the rarer medium at $C$ and $D$.
$\Delta= 2\mu t \: cos\:r $
Constructive Interference due to transmitted ray:
for constructive interference
$\Delta=n \lambda$
so from the above equations, we get
$2\mu t \: cos\: r =n \lambda$
Destructive Interference due to transmitted ray:
for destructive interference
$\Delta=(2n \pm 1)\frac{\lambda}{2}$
so from the above equations, we get
$2\mu t \: cos\: r =(2n \pm 1)\frac{\lambda}{2}$
$2\mu t \: cos\: r =n \lambda$
