Question

In Young's double slit experiment, the wavelength of light used is doubled and the distance between the two slits is halved, the new fringe width becomes,

• A. $3\text{times}$
• B. $\frac{1}{2}\text{times}$
• C. $4\text{times}$
• D. $2\text{times}$

Answer 1

The correct option is C $4\text{times}$

Fringe width is given by,
$\beta =\frac{\lambda D}{d}$

Here, $\lambda$ is the wavelength of light, $D$ is the distance between the slit and the screen and $d$ is the distance between the two slits.

According to the given information,
${\lambda }^{\prime }=2\lambda$, and $d^{\prime }=\frac{d}{2}$
$\therefore {\beta }^{\prime }=\frac{{\lambda }^{\prime }D}{d^{\prime }}=\frac{4\lambda D}{d}=4\beta$
So, the new fringe width becomes $4$ times.

Hence, $\left(D\right)$ is the correct answer.

$$\begin{array}{c}\text{Why this question:}\\ \text{Concept: Fringe width is inversely proportional to}\\ \text{the distance between the slits(d),}\text{and directly proportional to wavelength}\lambda .\end{array}$$