Question 143: Waves Optics

Question

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) :

Accepted Answer

As we know,
Path difference introduced by thin film,
$$
\Delta=(\mu-1) t
$$ .......(i)
and if fringe pattern shifts by one frings width, then path difference,
$$
\Delta=1 	imes \lambda=\lambda
$$ .......(ii)
So, from Eqs. (i) and (ii), we get
$$
(\mu-1) t=\lambda
$$
$$ \Rightarrow $$ $$
t=\frac{\lambda}{\mu-1}
$$

Answer

$${\lambda  \over {2\left( {\mu  - 1} \right)}}$$

Answer

$${\lambda  \over {\left( {2\mu  - 1} \right)}}$$

Answer

$${{2\lambda } \over {\left( {\mu  - 1} \right)}}$$

Answer

$${\lambda  \over {\left( {\mu  - 1} \right)}}$$

Waves Optics

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) :

Choose the correct answer: