Physics - Wave Optics Question with Solution | TestHub

The fringe width in a biprism experiment is given by $\beta =\frac{\lambda D}{d}$

Initial fringe width is ${\beta }_1=\frac{{\lambda }_1{D}_1}{d}$

Final fringe width is ${\beta }_2=\frac{{\lambda }_2{D}_2}{d}$

Taking the ratio of the two fringe widths:

$\frac{{\beta }_2}{{\beta }_1}=\frac{{\lambda }_2{D}_2}{{\lambda }_1{D}_1}$

Substituting the given values ${\lambda }_1=4800\text{ A˚}$, $D_1=1\text{ m}$, ${\lambda }_2=6400\text{ A˚}$, and $D_2=1+0.2=1.2\text{ m}$:

$\frac{{\beta }_2}{{\beta }_1}=\frac{6400\times 1.2}{4800\times 1}=\frac{4}{3}\times 1.2=1.6$

Percentage change in fringe width is given by:

$\frac{{\beta }_2-{\beta }_1}{{\beta }_1}\times 100=\left(\frac{{\beta }_2}{{\beta }_1}-1\right)\times 100$

$\Rightarrow (1.6-1)\times 100=60\%$

The fringe width increases by $60\%$.

Answer: increase by 60%.