In Young's double slits experiment, the position of 5 th bright...

Given here,

$D = 1m$, $\lambda = 600 \times {10}^{- 9}m$ and $n = 5$

Distance of nth bright fringe is given by, ${y}_{n} = \dfrac{n\lambdaD}{d}$, where, $d$ is separation between the slits.

$⇒ \dfrac{5 \times 600 \times {10}^{- 9} \times 1}{d} = 5 \times {10}^{- 2}$

$⇒ d = \dfrac{5 \times 600 \times {10}^{- 9} \times 1}{5 \times {10}^{- 2}} = 60 \times {10}^{- 6}m$

$⇒ d = 60\mum$.