Question

In Young's double slit experiment, the fringe width with light of wavelength $6000\mathring{A}$ is $3mm$. The fringe width, when the wavelength of light is changed to $4000\mathring{A}$ is

• A. $3mm$
• B. $1mm$
• C. $2mm$
• D. $4mm$

Answer 1

Answer: (C)

Fringe width $\beta =\frac{\lambda D}{d}$
where $\lambda$ is the wavelength of light, $D$ is distance between slits and the screen, $d$ is distance between the two slits.
As $D$ and $d$ remain the same, $\beta \propto \lambda$
$or\frac{{\beta }^{\prime }}{\beta }=\frac{{\lambda }^{\prime }}{\lambda }$ or
${\beta }^{\prime }=\frac{{\lambda }^{\prime }\beta }{\lambda }$
Substituting the given values, we get
${\beta }^{\prime }=\frac{4000\mathring{A}\times 3mm}{6000\mathring{A}}=2mm$