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Q. 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

KEAMKEAM 2007Wave Optics

Solution:

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