A parallel beam of light travelling in air (refractive index 1.0) is incident on a convex spherical glass surface of radius of curvature 50 cm. Refractive index of glass is 1.5. The rays converge to a point at a distance $$x$$ cm from the centre of the curvature of the spherical surface. The value of $$x$$ is ____ cm

A parallel beam of light (from infinity) in air is incident on a convex spherical glass surface (radius R = 50 cm, $$n_{\text{glass}} = 1.5$$). To find the distance $$x$$ from the center of curvature to the point of convergence, we apply the refraction formula at a spherical surface: $$\frac{n_2}{v} - \frac{n_1}{u} = \frac{n_2 - n_1}{R}$$ where $$n_1 = 1.0$$ (air), $$n_2 = 1.5$$ (glass), $$R = +50$$ cm (convex surface, center of curvature on the glass side), and $$u = -\infty$$ (parallel beam).

Substituting the values gives $$\frac{1.5}{v} - \frac{1}{-\infty} = \frac{1.5 - 1}{50}$$, so $$\frac{1.5}{v} = \frac{0.5}{50} = 0.01$$ and hence $$v = \frac{1.5}{0.01} = 150$$ cm. The image forms at 150 cm from the surface, inside the glass.

The center of curvature is at 50 cm from the surface (inside the glass), while the image is at 150 cm from the surface. Therefore, $$x = 150 - 50 = 100$$ cm.

The answer is 100 cm.