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Refraction at a Spherical Surfaces

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Estimated time: 12 minutes
CBSE: Class 12

Introduction

A spherical refracting surface is a curved boundary separating two transparent media, where one medium has refractive index n1​ and the other has refractive index n2.

When light travels from one medium to another through this curved surface, the rays bend due to refraction, and an image is formed depending on the curvature of the surface and the refractive indices of the two media.

This concept forms the basis for understanding lenses, because a thin lens may be treated as a combination of two spherical refracting surfaces.

CBSE: Class 12

Formula: Refraction at a Spherical Surface

For refraction at a spherical surface, the relation is:

\[\frac{n_2}{v}-\frac{n_1}{u}=\frac{n_2-n_1}{R}\]

CBSE: Class 12

Geometrical Setup

Consider an object point O placed on the principal axis of a spherical surface separating medium 1 and medium 2.

  • Medium 1 has a refractive index n1.
  • Medium 2 has a refractive index n2.
  • The centre of curvature is C.
  • The object distance is u.
  • The image distance is v.
  • The radius of curvature is R.

Step Action Significance
1 A light ray travels from an object to a spherical boundary. Establishes incident medium.
2 Ray strikes a curved surface and refracts. Bending depends on refractive indices.
3 Refracted rays meet or appear to meet. This determines the image position.
4 Formula links u, v, R, n₁, n₂. Used for solving numericals.
CBSE: Class 12

Cartesian Sign Convention

  • All distances are measured from the pole.
  • Distances measured in the direction of incident light are taken as positive.
  • Distances measured opposite to the direction of incident light are taken as negative.
  • Heights above the principal axis are positive.
  • Heights below the principal axis are negative.
Quantity Sign rule Typical case
u Usually negative for a real object placed to the left of the surface. Most board numericals
v Positive for real image on the right, negative for virtual image on the left. Depends on the image formed
R Positive if the centre of curvature lies on the positive side, negative otherwise. Depends on the surface shape
CBSE: Class 12

Conceptual Understanding

When Light Goes from a Rarer to a Denser Medium

The refracted ray bends towards the normal, and the image position depends on the curvature of the spherical surface and the object location.

When Light Goes from a Denser to a Rarer Medium

The refracted ray bends away from the normal, and the image may shift differently for the same object distance because the refractive index contrast changes.

Real-Life Analogy

Looking through a curved glass bowl or a transparent marble gives a distorted view of an object because light is refracted at a curved surface rather than a flat one.

CBSE: Class 12

Derivation Outline

Stepwise Logic

  1. Consider a paraxial ray from object O incident on the spherical surface.
  2. Draw the normal at the point of incidence by joining that point to the centre of curvature C.
  3. Use the small-angle approximation for the angles made by the incident ray, refracted ray, and normal.
  4. Apply the refraction relation in its small-angle form.
  5. Rearrange terms in terms of u, v, and R.
  6. Obtain the standard formula for refraction at a spherical surface.
CBSE: Class 12

Plane Surface vs Spherical Surface Refraction

Feature Plane Surface Spherical Surface
Boundary shape Flat interface Curved interface forming part of a sphere.
Image shift The apparent depth type effect is common Image location depends on curvature and refractive indices.
Main variables Refractive indices and depth u, v, R, n₁, n₂.
CBSE: Class 12

Example

An object is placed at a distance of 100 cm in air from a convex spherical surface of glass of refractive index 1.5. The radius of curvature is 20 cm. Find the image distance.

Given

  • n1 = 1 for air.
  • n2 = 1.5 for glass.
  • u = −100 cm.
  • R = +20 cm.

Formula

\[\frac{n_2}{v}-\frac{n_1}{u}=\frac{n_2-n_1}{R}\]

Substitution

\[ \frac{1.5}{v} - \frac{1}{100} = \frac{0.5}{20} \]

\[ \frac{1.5}{v} + \frac{1}{100} = \frac{1}{40} \]

\[ \frac{1.5}{v} = \frac{1}{40} - \frac{1}{100} = \frac{3}{200} \]

\[ v = 100\ \text{cm} \]

Result

The image is formed 100 cm from the spherical surface inside the glass medium.

Video Tutorials

We have provided more than 1 series of video tutorials for some topics to help you get a better understanding of the topic.

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