Question

In a certain double slit experimental arrangement, interference fringes of width 1 mm each are observed when light of wavelength $5000\mathring{A}$ is used. Keeping the setup unaltered, if the source is replaced by another of wavelength $6000\mathring{A}$, the fringe width will be

• A. 1.2 mm
• B. 0.5 mm
• C. 1 mm
• D. 1.5 mm

Answer 1

Answer: (A)

Fringe width, $\beta =\frac{\lambda D}{d}$
Since D and d are unaltered, $\beta \infty \lambda$
$\therefore \frac{{\beta }^{\prime }}{\beta }\frac{{\lambda }^{\prime }}{\lambda }$
or ${\beta }^{\prime }=\beta \times \frac{{\lambda }^{\prime }}{\lambda }=1\times \frac{6000}{5000}=1.2$ mm