Light from a point source in air falls on a spherical glass surface (refractive index, $$\mu = 1.5$$ and radius of curvature = 50 cm). The image is formed at a distance of 200 cm inside the glass. The magnitude of distance of the light source from the glass surface is ________ m.

Take the pole of the spherical surface as the origin and measure all distances from this point along the principal axis (Cartesian sign convention).

• Incident medium (air): $$n_1 = 1$$.
• Refracting medium (glass): $$n_2 = 1.5$$.
• Radius of curvature: the centre of curvature lies inside the glass, i.e. on the side to which light travels, so $$R = +50\,\text{cm}$$.
• Image is formed inside the glass at a distance $$v = +200\,\text{cm}$$ (positive because it is on the side of the refracted light).

For refraction at a spherical surface, the Cartesian-form equation is

$$\frac{n_2}{v} - \frac{n_1}{u} = \frac{n_2 - n_1}{R} \quad -(1)$$

Substitute the given data into $$(1)$$:

$$\frac{1.5}{200} - \frac{1}{u} = \frac{1.5 - 1}{50}$$

Simplify each term:
$$\frac{1.5}{200} = 0.0075$$,
$$\frac{1.5 - 1}{50} = \frac{0.5}{50} = 0.01$$.

Hence,

$$0.0075 - \frac{1}{u} = 0.01$$

Rearrange to solve for $$\frac{1}{u}$$:

$$-\frac{1}{u} = 0.01 - 0.0075 = 0.0025$$

Therefore,

$$\frac{1}{u} = -0.0025 \;\Rightarrow\; u = -\frac{1}{0.0025} = -400\,\text{cm}$$

The negative sign confirms that the object (light source) is on the side opposite to the direction of positive measurement, i.e. in air in front of the surface.

Magnitude of the object distance:
$$|u| = 400\,\text{cm} = 4\,\text{m}$$.

Hence, the light source is $$4\,\text{m}$$ in front of the glass surface.