A young’s double-slit experiment is performed using monochromatic light of wavelength $\lambda$. The intensity of light at a point on the screen, where the path difference is $\lambda$, is K units. The intensity of light at a point where the path difference is $\frac{\lambda }{6}$is given by $\frac{\mathrm{nK}}{12}$, where n is an integer. The value of n is ______ 

In Youngs double slit experiment, intensity at a point is given by 

$I=I_0{\cos }^2\frac{\varphi }{2}$…….(1)

Where ,$\varphi =$ phase difference,

Using phase difference, $\varphi =\frac{2\pi }{\lambda }\times$path difference

For path difference $\lambda$, phase difference ${\varphi }_1=2\pi$

For path difference ,$\frac{\lambda }{6}$ , phase difference ${\varphi }_2=\frac{\pi }{3}$

Using equation (i), 

$$\begin{gathered} \frac{I_1}{I_2}=\frac{{\cos }^2\left(\frac{{\varphi }_1}{2}\right)}{{\cos }^2\left(\frac{{\varphi }_2}{2}\right)}=\frac{{\cos }^2\left(\frac{2\pi }{2}\right)}{{\cos }^2\left(\frac{\pi }{6}\right)}\Rightarrow \frac{K}{I_2}=\frac{1}{{\left({\displaystyle \frac{\sqrt{3}}{2}}\right)}^2}=\frac{4}{3} \\ \Rightarrow I_2=\frac{3K}{4}=\frac{9\mathrm{K}}{12}\therefore \mathrm{n}=9. \end{gathered}$$