Derivation of Fringe width of the wedge-shaped thin film:
The distance between two consecutive bright (or dark) fringes is called the fringe width.
If the $n^{th}$ bright fringe is formed at a distance $x_{n}$ from the edge of the wedge shaped film where the thickness is $t_{n}$. So the path difference for $x_{n}$ bright fringe:
$(2n-1) \frac{\lambda}{2}= 2 \mu \: t_{n} \: cos(\alpha+r) \qquad(1)$
For Normal IncidenceThe incident Angle $i=0$The refracted Angle $r=0$
then from equation $(1)$
$(2n-1) \frac{\lambda}{2}= 2 \mu \: t_{n} \: cos\alpha \qquad(2)$
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| Fringe width of wedge-shaped thin film for normal incidence |
From the figure, In $\Delta OAB$
$tan \alpha = \frac{t_{n}}{x_{n}}$
$t_{n}=x_{n} \: tan \alpha \qquad(3) $
Now put the value of $t_{n}$ in equation $(2)$
$(2n-1) \frac{\lambda}{2}= 2 \mu \: x_{n} \: tan \alpha \: cos\alpha $
$(2n-1) \frac{\lambda}{2}= 2 \mu \: x_{n}\: \frac{sin \alpha}{cos\alpha} \: cos\alpha $
$(2n-1) \frac{\lambda}{2}= 2 \mu \: x_{n} \: sin \alpha \qquad(4)$
Similarly for $(n+1)^{th}$ bright fringe i.e put the $(n+1)$ in place of $n$ in above equation then we get
$\left\{ 2(n+1)-1 \right\} \frac{\lambda}{2}= 2 \mu \: x_{n+1} \: sin \alpha $
$(2n+1) \frac{\lambda}{2}= 2 \mu \: x_{n+1} \: sin \alpha \qquad(5)$
Subtract the equation $(5)$ and equation$(4)$
$2\mu \: (x_{n+1}-x_{n})\: sin \alpha =\lambda$
$2\mu \beta \: sin \alpha =\lambda$
Where $\beta=$ fringe Width$(x_{n+1}-x_{n})$
$\beta=\frac{\lambda}{2\mu\: sin \alpha}$
If $\alpha$ is very small so $sin \alpha \approx \alpha$
$\beta=\frac{\lambda}{2\mu\: \alpha}$
