Optics Ques 4
- In a double slit experiment, when a thin film of thickness $t$ having refractive index $\mu$ is introduced in front of one of the slits, the maximum at the centre of the fringe pattern shifts by one fringe width. The value of $t$ is $(\lambda$ is the wavelength of the light used)
(2019 Main, 12 April I)
(a) $\frac{2 \lambda}{(\mu-1)}$
(b) $\frac{\lambda}{2(\mu-1)}$
(c) $\frac{\lambda}{(\mu-1)}$
(d) $\frac{\lambda}{(2 \mu-1)}$
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Answer:
Correct Answer: 4.( c )
Solution:
Formula:
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} $
BETA
