Using Young's double slit experiment, a monochromatic light of...

We know that fringe width is given by, $\beta = \dfrac{\lambdaD}{d}$

$⇒ \dfrac{{\beta}^{'}}{\beta} = \dfrac{{\lambda}^{'}}{\lambda} \times \dfrac{d}{{d}^{'}} = \dfrac{6000}{5000} \times \dfrac{d}{2d}$

$⇒ {\beta}^{'} = \dfrac{6}{5 \times 2} \times 0 . 5$$= 0 . 3mm$