In a double slit experiment, the distance between slits is increased 10 times whereas their distance from screen is halved, then what is the fringe width?

The correct option is: become 1/20.

In Young's Double Slit Experiment, the fringe width, denoted by:
$\beta$
is the separation between two consecutive bright or dark fringes on the screen. The formula for the initial fringe width, denoted by:
${\beta }_1$
is given by:

$${\beta }_1=\frac{\lambda D_1}{d_1}$$

where:
•
$\lambda$
is the wavelength of the light source,
•
$D_1$
is the initial distance between the slits and the screen, and
•
$d_1$
is the initial distance between the two slits.

The problem states that the experimental setup is modified as follows:
• The distance between the slits is increased by 10 times:
$d_2=10d_1$
• The distance from the screen is halved:
$D_2=\frac{D_1}{2}$

We can write the expression for the new fringe width, denoted by:
${\beta }_2$
as follows:

$${\beta }_2=\frac{\lambda D_2}{d_2}$$

Now, substitute the new values of
$d_2$
and
$D_2$
into the equation:

$${\beta }_2=\frac{\lambda \left(\frac{D_1}{2}\right)}{10d_1}$$

We can factor out the constants to relate this back to the original fringe width:

$${\beta }_2=\frac{1}{2\times 10}\times \frac{\lambda D_1}{d_1}$$

$${\beta }_2=\frac{1}{20}{\beta }_1$$

Thus, the fringe width becomes:
$\frac{1}{20}$
of its original value.