In Young's double slit experiment, the intensity of light at a point on the screen where the path difference is $\lambda$ is $\mathrm{K}$ units ($\lambda$ is the wavelength of light used). The percentage change in intensity at a point where the path difference is $\dfrac{\lambda}{6}$ and the above point is

The intensity $I$ at any point on the screen in Young's double slit experiment is given by $I = I_{max} \cos^{2}\left(\dfrac{\phi}{2}\right)$, where $\phi$ is the phase difference.

The phase difference $\phi$ is related to the path difference $\Delta x$ by $\phi = \dfrac{2\pi}{\lambda} \Delta x$.

For the first point, the path difference is $\Delta x_1 = \lambda$. Thus, $\phi_1 = \dfrac{2\pi}{\lambda} \times \lambda = 2\pi$.

The intensity at this point is $I_1 = I_{max} \cos^{2}\left(\dfrac{2\pi}{2}\right) = I_{max} \cos^{2}(\pi) = I_{max}(1)^2 = I_{max}$. Given $I_1 = K$, we have $I_{max} = K$.

For the second point, the path difference is $\Delta x_2 = \dfrac{\lambda}{6}$. Thus, $\phi_2 = \dfrac{2\pi}{\lambda} \times \dfrac{\lambda}{6} = \dfrac{\pi}{3}$.

The intensity at this point is $I_2 = I_{max} \cos^{2}\left(\dfrac{\pi/3}{2}\right) = I_{max} \cos^{2}\left(\dfrac{\pi}{6}\right) = K \left(\dfrac{\sqrt{3}}{2}\right)^2 = K \times \dfrac{3}{4} = 0.75K$.

The change in intensity is $I_1 - I_2 = K - 0.75K = 0.25K$.

The percentage change in intensity is $\dfrac{I_1 - I_2}{I_1} \times 100 = \dfrac{0.25K}{K} \times 100 = 25\%$.

Answer: 25%