Question

young's double slit experiment while using a source of light of wavelength 5000 am strong the fringe width is 0.6 CM if the distance between the screen and the slit is reduced by the half of what should be the wavelength of the height to get the fringe width of 0.4cm​

Answer 1

Answer:

hi can you use atmagyan

Explanation:

Mark my answer in brainlist please

Answer 2

$\huge\color{blue}{ \colorbox{pink}{\colorbox{white} {Solution}}}$

Initial fringe width    βi=0.6 cm=0.006 m

Final fringe width  βf=0.003=βi/2

Fringe width    β=dλD

⟹ βf/βi=λf Df/λi Di

Or    1/2=λf (Di/2)/5000×Di

⟹ λf=5000 A⁰