In Young’s double-slit experiment, the ${8^{th}}$ maximum with wavelength ${\lambda _1}$ is at distance ${d_1}$ from the central maximum and the ${6^{th}}$ maximum with wavelength \[{\lambda _2}\]is at distance ${d_2}$ from the central maximum. Then ${d_1}/{d_2}$ is equal to :
A) $\dfrac{{{\lambda _1}}}{{{\lambda _2}}}$
B) $\dfrac{{4{\lambda _1}}}{{3{\lambda _2}}}$
C) $\dfrac{{3{\lambda _2}}}{{4{\lambda _1}}}$
D) $\dfrac{{3{\lambda _1}}}{{4{\lambda _2}}}$

Hint: To calculate the ratio, we need to find the expression for the fringes at ${8^{th}}$ and ${6^{th}}$ maxima respectively. By taking the ratio of the fringes, we can obtain the relation between wavelength and distance from the central maxima.

Formula used:
${x_n} = \dfrac{{n\lambda D}}{d}$
where:
${x_n} = $ nth bright fringe
$n = $ No. of the fringe
$D = $Distance between the screen and the slits.
$d = $Distance from the central maxima
$\lambda = $Wavelength of the light

Complete step by step solution:
We know, when two light waves superimpose on each other, a resultant wave having an amplitude less than or greater than the amplitude of the original wave. This phenomenon is known as interference.
In Young's double slit experiment, we know it was concluded that when the light in waveform passes through two slips they interfere with each other giving rise to alternate dark and bright fringes. Dark fringes were formed at the location of destructive interference and bright fringes were formed where the waves interfere constructively. Bright fringes were formed at the maxima whereas dark fringes were formed at minima.
We also came to know about the formula for bright fringes, this was given by:
${x_n} = \dfrac{{n\lambda D}}{d}$
where:
${x_n} = $ nth bright fringe
$n = $ No. of the fringe
$D = $Distance between the screen and the slits.
$d = $Distance from the central maxima
$\lambda = $Wavelength of the light
Therefore, for ${8^{th}}$maxima, we can write:
$\Rightarrow {x_8} = \dfrac{{8{\lambda _1}D}}{{{d_1}}}$
On rearranging the equation, we obtain:
$\Rightarrow {d_1} = \dfrac{{6{\lambda _2}D}}{{{x_8}}}$
And, for ${6^{th}}$ maxima, we can write:
$\Rightarrow {x_6} = \dfrac{{6{\lambda _2}D}}{{{d_2}}}$
On rearranging the equation, we obtain:
$\Rightarrow {d_2} = \dfrac{{6{\lambda _2}D}}{{{x_6}}}$
Now, we take the ratio of ${d_1}/{d_2}$, thus we obtain:
$\Rightarrow \dfrac{{{d_1}}}{{{d_2}}} = \dfrac{{{x_8}}}{{{x_6}}} = \dfrac{{8{\lambda _1}}}{{6{\lambda _2}}}$
On reducing to the smallest terms, we arrive at:
$\Rightarrow \dfrac{{{d_1}}}{{{d_2}}} = \dfrac{{4{\lambda _1}}}{{3{\lambda _2}}}$.

Therefore, option (B) is correct.

Note: For interference to take place, the necessary conditions are that the sources must be coherent, which means they must emit identical waves and they should have an equal phase difference or constant phase difference, and that they must be monochromatic in nature.