Question

Two wavelengths ${\lambda }_1$ and ${\lambda }_2$ are used in bi-prism experiment. If ${\lambda }_1$ is 430 nm, the value of the wavelength ${\lambda }_2$ such that its fourth order bright fringe falls over the sixth order bright fringe of ${\lambda }_1$ would be

• A. $645nm$
• B. $64.5nm$
• C. $6.45nm$
• D. $0.645nm$

Answer 1

The correct option is A $645nm$

Fourth order bright fringe of ${\lambda }_2=$ sixth order bright fringe of ${\lambda }_1$
$\frac{4D{\lambda }_2}{d}=\frac{6D{\lambda }_1}{d}$
$4{\lambda }_2=6{\lambda }_1$
${\lambda }_2=\frac{6}{4}{\lambda }_1$
$=\frac{6}{4}\left(430\right)$
$=645nm$