Question

State two conditions to obtain sustained interference of light. In Young's double slit experiment. using light of wavelength $400nm$ , interference fringes of width ${}^{\prime }{X}^{\prime }$ are obtained. The wavelength of light is increased to $600nm$ and the separation between the slits is halved. If one wants the observed fringe width on the screen , to be the same in the two cases, find the ratio of the distance between the screen and the plane of the slits in the two arrangements.

Answer 1

We have,

${\beta }_1=\frac{400\times {10}^{-9}\times D_1}{d_1}$

${\beta }_2=\frac{600\times {10}^{-9}\times D_2}{d_1/2}$

Since ${\beta }_1={\beta }_2$

So diving equation second by first, we get

$\frac{{\beta }_2}{{\beta }_1}=\frac{600\times {10}^{-9}{D}_2\times 2d_1}{d_1\times 400\times {10}^{-9}\times D_1}$

$1=\frac{1200D_2}{400D_1}$

$D_1=3D_2$

$\frac{D_2}{D_1}=\frac{1}{3}$