Question

In Young's double slit experiment, when violet light of wavelength $4358\, \mathring{A}$ is used, the $84$ fringes are seen in the field of view, but when sodium light of certain wavelength is used, then $62$ fringes are seen in the field of view, the wavelength of sodium light is

• A. $6893\, \mathring{A}$
• B. $5904\, \mathring{A}$
• C. $5523\, \mathring{A}$
• D. $6429\, \mathring{A}$

Answer 1

Answer: (B)

As field of view is same in both cases
$n_{1} \beta_{1} =n_{2} \beta_{2}$
or $n_{1}\left(\frac{D \lambda_{1}}{d}\right) =n_{2}\left(\frac{D \lambda_{2}}{d}\right)$
or $\lambda_{2}=\left(\frac{n_{1}}{n_{2}}\right) \lambda_{1}$
$\therefore \lambda_{2} =\left(\frac{84}{62}\right) \times 4358$
$\lambda_{2} =5904\, \mathring{A}$