Question

In the Young's experiment, using light of $\lambda =5893A^0$,62 fringes are observed in the fields of view.how many fringers will be formed in the field of view for wavelength of $\lambda =5461A^0$:-

• A. $62$
• B. $66$
• C. $60$
• D. $58$

Answer 1

The correct option is D $58$

Given,

${\beta }_1=62$

${\lambda }_1=5893A^0$

$\beta =?$

${\lambda }_2=5461A^0$

Fringe width in young's experiment,

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

$\beta \propto \lambda$

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

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

${\beta }_2=62\times \frac{5461A^0}{5893A^0}$

${\beta }_2=58$