Question

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

The correct option is D $84$

Given, ${\lambda }_1=5893\stackrel{\circ }{A},n_1=62,{\lambda }_2=4358\stackrel{\circ }{A}$
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 $n_2=n_1\left(\frac{{\lambda }_1}{{\lambda }_2}\right)$
$=62\times \frac{5893}{4358}=84$