Question

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

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

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, $\beta \propto \lambda$
$or \frac{\beta'}{\beta} =\frac{\lambda'}{\lambda}$ or
$\beta' = \frac{\lambda'\beta}{\lambda}$
Substituting the given values, we get
$\beta' = \frac{4000\,\mathring{A}\times 3\,mm}{6000 \,\mathring{A}} = 2\,mm $