Two transparent media having refractive indices 1.0 and 1.5 are separated by a spherical refracting surface of radius of curvature 30 cm. The centre of curvature of surface is towards denser medium and a point object is placed on the principal axis in rarer medium at a distance of 15 cm from the pole of the surface. The distance of image from the pole of the surface is _______ cm.

We have $$\mu_1 = 1.0$$ (rarer medium), $$\mu_2 = 1.5$$ (denser medium), radius of curvature $$R = 30$$ cm (centre of curvature towards the denser medium, so $$R = +30$$ cm), and object distance $$u = -15$$ cm (object in the rarer medium).

Using the refraction formula at a spherical surface:

$$\frac{\mu_2}{v} - \frac{\mu_1}{u} = \frac{\mu_2 - \mu_1}{R}$$

Substituting the values:

$$\frac{1.5}{v} - \frac{1.0}{-15} = \frac{1.5 - 1.0}{30}$$

$$\frac{1.5}{v} + \frac{1}{15} = \frac{0.5}{30}$$

$$\frac{1.5}{v} + \frac{1}{15} = \frac{1}{60}$$

$$\frac{1.5}{v} = \frac{1}{60} - \frac{1}{15} = \frac{1}{60} - \frac{4}{60} = \frac{-3}{60} = \frac{-1}{20}$$

$$v = 1.5 \times (-20) = -30 \text{ cm}$$

The negative sign indicates the image is formed on the same side as the object (in the rarer medium), at a distance of 30 cm from the pole. So, the answer is $$30$$ cm.