Question

Light from a source emitting two wavelengths ${\lambda }_1$ and ${\lambda }_2$ is allowed to fall on Young’s double slit apparatus after filtering one of the wavelengths. The position of interference maxima is noted. If the 4th maxima due to ${\lambda }_1$ and the third maxima due to ${\lambda }_2$ coincide while the other wavelength is filtered in respective cases, then $\frac{{\lambda }_2}{{\lambda }_1}=$. (Answer upto 2 decimal digits)

Answer 1

The maxima in this case is obtained whenever
$x=\frac{n\lambda }{d}$
$\therefore x_4=\frac{4{\lambda }_1}{d}$
Similarly
$x_3=\frac{3{\lambda }_2}{d}$
$\therefore \frac{4{\lambda }_1}{d}=\frac{3{\lambda }_2}{d}$
$\Rightarrow \frac{{\lambda }_2}{{\lambda }_1}=1.33$