JEE MAIN - Physics (2019 - 12th April Morning Slot - No. 14)
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) :
$${\lambda \over {2\left( {\mu - 1} \right)}}$$
$${\lambda \over {\left( {2\mu - 1} \right)}}$$
$${{2\lambda } \over {\left( {\mu - 1} \right)}}$$
$${\lambda \over {\left( {\mu - 1} \right)}}$$
Explanation
As we know,
Path difference introduced by thin film,
$$ \Delta=(\mu-1) t $$ .......(i)
_12th_April_Morning_Slot_en_14_1.png)
and if fringe pattern shifts by one frings width, then path difference,
$$ \Delta=1 \times \lambda=\lambda $$ .......(ii)
So, from Eqs. (i) and (ii), we get
$$ (\mu-1) t=\lambda $$
$$ \Rightarrow $$ $$ t=\frac{\lambda}{\mu-1} $$
$$ \Delta=(\mu-1) t $$ .......(i)
and if fringe pattern shifts by one frings width, then path difference,
$$ \Delta=1 \times \lambda=\lambda $$ .......(ii)
So, from Eqs. (i) and (ii), we get
$$ (\mu-1) t=\lambda $$
$$ \Rightarrow $$ $$ t=\frac{\lambda}{\mu-1} $$
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