Question

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 from the glass surface inside the glass. The magnitude of distance of the light source from the glass surface is ____________________ m.

Answer 1

4 m

Answer 2

We use the formula for refraction at a spherical surface:

$$\frac{n'}{v} - \frac{n}{u} = \frac{n' - n}{R}$$

Here:

• $n = 1$ (air)
• $n' = 1.5$ (glass)
• $v = 200\text{ cm} = 2\text{ m}$ (image distance inside the glass)
• $R = 50\text{ cm} = 0.5\text{ m}$

Substitute the values:

$$\frac{1.5}{2} - \frac{1}{u} = \frac{1.5 - 1}{0.5}$$

Simplify the right-hand side:

$$\frac{1.5 - 1}{0.5} = \frac{0.5}{0.5} = 1$$

Now, solve for $\frac{1}{u}$:

$$\frac{1.5}{2} - \frac{1}{u} = 1 \quad \Rightarrow \quad 0.75 - \frac{1}{u} = 1$$ $$-\frac{1}{u} = 1 - 0.75 = 0.25 \quad \Rightarrow \quad \frac{1}{u} = -0.25$$ $$u = -4\text{ m}$$

The negative sign indicates that the object is located on the same side as the incident light. However, since the question asks for the magnitude, the answer is 4 m.