Question

While conducting Young's double slit experiment it was noted that fringe width with light of wavelength $6000 \, \mathring{A} $ was $3 \, mm$ . If the experiment is repeated but this time wavelength of light used is $4000 \, \mathring{A} $ , what will be the new fringe width?

• 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
$\beta ^{'}=\frac{4000 Å \times 3 \, m m}{6000 Å}=2 \, mm$