Question

In Young's experiment when sodium light of wavelength 5893 $\mathring{A}$ is used, then 62 fringes are seen in the field of view. Instead, if violet light of wavelength 4358 $\mathring{A}$ is used, then the number of fringes that will be seen in the field of view will be

• A. 54
• B. 64
• C. 74
• D. 84

Answer 1

Answer: (D)

Given ${\lambda }_1=5893\mathring{A},n_1=62,\lambda =4358\mathring{A}$
As field of view in case of both the wavelengths is same,
hence
$n_1{\beta }_1=n_2{\beta }_2$
as $n_1\left(\frac{D{\lambda }_1}{d}\right)=n_2\left(\frac{D{\lambda }_2}{d}\right)$
or $n_2=n_1\left(\frac{{\lambda }_1}{{\lambda }_2}\right)$
$=62\times \frac{5893}{4358}=84$