CBSE 2025, Set 1

Question

A point object is placed in air at a distance $R/3$ in front of a convex surface of radius of curvature $R$, separating air from a medium of refractive index $n$ ($< 4$). Find the nature and position of the image formed.

Edvaya · Accepted Answer

The correct answer is Subjective. Using the refraction formula at a spherical surface: $$\frac{n_2}{v} - \frac{n_1}{u} = \frac{n_2 - n_1}{R}$$ Here: $n_1 = 1$ (air), $n_2 = n$ (medium), $u = -R/3$ (object is in air, to the left) The radius of curvature R is positive for a convex surface (centre of curvature on transmission side). Substituting: $$\frac{n}{v} - \frac{1}{(-R/3)} = \frac{n - 1}{R}$$ $$\frac{n}{v} + \frac{3}{R} = \frac{n - 1}{R}$$ $$\frac{n}{v} = \frac{n - 1}{R} - \frac{3}{R} = \frac{n - 4}{R}$$ $$v = \frac{nR}{n - 4}$$ Since $n **Interpretation:** A virtual, erect, and diminished image is formed on the same side as the object. This occurs because the object is placed between the surface and its first focal point.

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Physics > Optics > Ray Optics And Optical Instruments

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