Physics - Wave Optics Question with Solution | TestHub

The distance of the $n$-th dark fringe from the central bright fringe in an interference pattern is given by:

$x_{nD}=(2n-1)\frac{\lambda D}{2d}$

For the second dark fringe ($n=2$):

$x_{2D}=(2(2)-1)\frac{\lambda D}{2d}=\frac{3}{2}\frac{\lambda D}{d}$

Given that $x_{2D}=3$ mm:

$\frac{3}{2}\frac{\lambda D}{d}=3$

$\frac{\lambda D}{d}=2$ mm

The distance of the $n$-th bright fringe from the central fringe is given by:

$x_{nB}=n\frac{\lambda D}{d}$

For the sixth bright fringe ($n=6$):

$x_{6B}=6\times 2=12$ mm

Answer: 12 mm