Physics - Wave Optics Question with Solution | TestHub

The initial fringe width is given by ${\beta }_1=\frac{{\lambda }_1{D}_1}{d}$

The final fringe width is given by ${\beta }_2=\frac{{\lambda }_2{D}_2}{d}$

It is given that the fringe width is reduced by $20\%$ of its initial value, so:

${\beta }_2={\beta }_1-0.2{\beta }_1=0.8{\beta }_1$

Substituting the expressions for fringe widths:

$\frac{{\lambda }_2{D}_2}{d}=0.8\frac{{\lambda }_1{D}_1}{d}$

$D_2=\frac{0.8{\lambda }_1{D}_1}{{\lambda }_2}$

Given ${\lambda }_1=5000$ \AA, ${\lambda }_2=6400$ \AA, and $D_1=100$ cm:

$D_2=\frac{0.8\times 5000\times 100}{6400}$

$D_2=\frac{400000}{6400}=62.5$ cm

The displacement of the screen is $\Delta D=D_1-D_2=100-62.5=37.5$ cm.

Since $D_2<D_1$, the screen should be moved towards the source by $37.5$ cm.

Answer: towards the source by $37.5$ cm