Question

A double slit interference experiment is carried out in air and the entire arrangement is dipped in water. As a result

• A. The fringe width decreases.
• B. The fringe width increases.
• C. The fringe width remains unchanged.
• D. Fringe pattern disappears.

Answer 1

The correct option is A The fringe width decreases.

When the $YDSE$ setup is dipped in any medium, then wavelength of light wave changes.

From Snell's law:
$\frac{{\mu }_2}{{\mu }_1}=\frac{{\lambda }_1}{{\lambda }_2}$

$\Rightarrow \frac{{\mu }_{water}}{{\mu }_{air}}=\frac{{\lambda }_a}{{\lambda }_w}$

$\Rightarrow {\lambda }_w=\frac{{\lambda }_a}{{\mu }_w}[\because {\mu }_{air}=1]$

Fringe width is given by,

$\beta =\frac{\lambda D}{d}$

Thus, ${\beta }_w=\frac{{\lambda }_{w}D}{d}$

$\Rightarrow {\beta }_w=\frac{{\lambda }_{a}D}{{\mu }_{w}d}$

$\Rightarrow {\beta }_w=\frac{{\beta }_{air}}{{\mu }_w}$

$\because {\mu }_w>{\mu }_a$

Thus, ${\beta }_w<{\beta }_a$

Fringe width decreases when the $YDSE$ setup is dipped in water.

Hence, option $\left(A\right)$ is correct.

Why this Question ? The position of central maxima is unaffected by the dipping of $YDSE$ setup in liquid. Because the path differences between rays from both slits will be zero at the same position.