CBSE 2025, Set 2

Question

(b) A point source of light in air is kept at a distance of $12	ext{ cm}$ in front of a convex spherical surface of glass of refractive index $1.5$ and radius of curvature $30	ext{ cm}$. Find the nature and position of the image formed.

Edvaya · Accepted Answer

The correct answer is Subjective. Given:
* Refractive index of first medium (air), $n_1 = 1$
* Refractive index of second medium (glass), $n_2 = 1.5$
* Object distance, $u = -12	ext{ cm}$ (using sign convention)
* Radius of curvature of convex surface, $R = +30	ext{ cm}$

The formula for refraction at a single spherical surface is:
$$\frac{n_2}{v} - \frac{n_1}{u} = \frac{n_2 - n_1}{R}$$

Substitute the given values:
$$\frac{1.5}{v} - \frac{1}{(-12)} = \frac{1.5 - 1}{30}$$
$$\frac{1.5}{v} + \frac{1}{12} = \frac{0.5}{30}$$
$$\frac{1.5}{v} + \frac{1}{12} = \frac{1}{60}$$
$$\frac{1.5}{v} = \frac{1}{60} - \frac{1}{12}$$

Take the LCM (which is 60):
$$\frac{1.5}{v} = \frac{1 - 5}{60}$$
$$\frac{1.5}{v} = \frac{-4}{60} = -\frac{1}{15}$$
$$v = -15 	imes 1.5$$
$$v = \boxed{-22.5	ext{ cm}}$$

**Position:** The image is formed at a distance of $22.5	ext{ cm}$ from the pole, on the same side as the object.
**Nature:** Since $v$ is negative, the refracted rays do not actually meet. Therefore, the image is **virtual and erect**.

Edvaya Target

Edvaya Aspire

Take a Free Mock Test

Target Subjects

Aspire Subjects

Subject Mastery

Physics > Optics > Ray Optics And Optical Instruments

Keep Practicing

Current Question

Community Stats

Track Your Progress

Edvaya Target

We Value Your Privacy