The image of an object placed in air formed by a convex refracting surface is at a distance of 10 m behind the surface. The image is real and is at $$\frac{2rd}{3}$$ of the distance of the object from the surface. The wavelength of light inside the surface is $$\frac{2}{3}$$ times the wavelength in air. The radius of the curved surface is $$\frac{x}{13}$$ m, the value of $$x$$ is ________.

The wavelength inside the surface is $$\frac{2}{3}$$ times the wavelength in air, which means the refractive index of the medium is $$\mu = \frac{\lambda_{\text{air}}}{\lambda_{\text{medium}}} = \frac{3}{2}$$.

The image is real and located 10 m behind the surface (in the denser medium), so $$v = +10$$ m. The image distance is $$\frac{2}{3}$$ of the object distance, meaning $$v = \frac{2}{3}|u|$$, so $$|u| = \frac{3}{2} \times 10 = 15$$ m and $$u = -15$$ m (object is on the same side as the incoming light).

Using the refraction formula for a single spherical surface, $$\frac{\mu_2}{v} - \frac{\mu_1}{u} = \frac{\mu_2 - \mu_1}{R}$$, where $$\mu_1 = 1$$ (air), $$\mu_2 = \frac{3}{2}$$, $$v = 10$$, $$u = -15$$, and $$R$$ is positive for a convex surface.

Substituting: $$\frac{3/2}{10} - \frac{1}{-15} = \frac{3/2 - 1}{R}$$, which gives $$\frac{3}{20} + \frac{1}{15} = \frac{1}{2R}$$. Computing the left side: $$\frac{3}{20} + \frac{1}{15} = \frac{9}{60} + \frac{4}{60} = \frac{13}{60}$$.

So $$\frac{13}{60} = \frac{1}{2R}$$, giving $$R = \frac{60}{26} = \frac{30}{13}$$ m. Since $$R = \frac{x}{13}$$, we get $$x = 30$$.