Question

In Young's double slit experiment, while using a source of light of wavelength 5000 A. The fringe width obtained is 0.6 cm. If the distance between the screen and the slit is, reduced to half, what should be the wavelength of the source to get fringes 0.003m wide?

Answer 1

Initial fringe width ${\beta }_i=0.6cm=0.006m$

Final fringe width ${\beta }_f=0.003=\frac{{\beta }_i}{2}$

Fringe width $\beta =\frac{\lambda D}{d}$

$⟹\frac{{\beta }_f}{{\beta }_i}=\frac{{\lambda }_f{D}_f}{{\lambda }_i{D}_i}$

Or $\frac{1}{2}=\frac{{\lambda }_f\left(D_i/2\right)}{5000\times D_i}$

$⟹{\lambda }_f=5000A^o$