Why does refraction occur at a spherical surface, and how is it different from refraction at a plane surface?

Refraction at a spherical surface occurs due to a change in light speed between media with different refractive indices. Unlike flat surfaces, the curvature causes light rays to converge or diverge, depending on whether the surface is convex or concave. This curvature also creates varying angles of incidence, leading to greater deviation in the light’s path.

Why is it that in spherical refraction, magnification may not be uniform in all directions for large objects?

For large objects, varying distances from the curved surface cause non-uniform magnification due to changes in object distance. In contrast, small objects under the paraxial approximation are nearly equidistant, resulting in uniform magnification.

How does spherical aberration impact the accuracy of the refraction formula?

The refraction formula uses the paraxial approximation, valid only for rays near the principal axis. Rays farther away deviate more and don't meet at the same point, causing spherical aberration and image blur. To reduce this, lenses often use aspherical surfaces or corrective elements.

Can a spherical refracting surface act as both converging and diverging under different conditions?

A single spherical surface can converge or diverge light depending on the direction of travel and refractive indices. For example, a convex surface converges light from a rarer to a denser medium, but diverges it in reverse. The behavior depends on both curvature and medium, not just shape.

Why is the concept of “optical center” not applicable to single spherical refracting surfaces?

The optical center exists only in lenses with two spherical surfaces. A single curved surface lacks symmetry and can't ensure rays pass undeviated through a fixed point. Thus, the concept doesn't apply to a single refracting surface.

New Launch

Boost your JEE prep with Major Test Series

Refraction at Spherical Surfaces

Refraction at a spherical surface happens when light passes from one transparent medium to another through a curved, spherical boundary. This bending of light changes the direction and position of the image formed. Unlike refraction at a flat surface, the curvature causes the light rays to converge or diverge, creating real or virtual images.This phenomenon is important in the functioning of lenses and many optical instruments. By studying refraction at spherical surfaces, we can understand how images are formed and learn to calculate the position and size of these images using the laws of optics.

1.0Law of Refraction At Spherical Surface

When light passes from a medium of refractive index $μ_{1}$ to a medium of refractive index 2 by a spherical surface of radius of curvature R then the relation between object distance u and image distance v is given by

$\frac{μ _{2}}{v} - \frac{μ _{1}}{u} = \frac{μ _{2} - μ _{1}}{R}$

2.0Terms Related To Refraction At Spherical Surfaces

(1) Centre of curvature (C)-It is the centre of a sphere of which the surface is a part.

(2) Radius of curvature (R)-It is the radius of the sphere of which the surface is a part.

(3) Pole (P)-It is the geometrical centre of the spherical refracting surface.

(4) Principal Axis-It is the straight line joining the centre of curvature to the pole.

(5) Focus (F)-When a narrow beam of rays of light, parallel to the principal axis and close to it (known as paraxial rays) is incident on a refracting spherical boundary, the refracted beam is found to converge or appear to diverge [depending upon the nature of the boundary (concave/convex) and the refractive index of two media] from a point on the principal axis. This point is called focus.

3.0Cartesian Sign Convention

(1) All distances are measured from the pole (P).

(2) Distances measured in the direction of incident rays are taken as positive.

(3) Distances above the principal axis are taken as positive.

4.0Derivation of Spherical Refraction

Suppose AB is the spherical surface which separates the two medium of refractive index $μ_{1}$ and $μ_{2}$
O and I are the position of object and image in the medium of R.I. $μ_{1}$ and $μ_{2}$

Respectively.

From $△ MOC \Rightarrow i = α + β$

$Δ MC I \Rightarrow β = r + θ$

For small aperture, i and r are small

from Snell’s law,

$μ_{1} sin i = μ_{2} sin r μ_{1} (α + β) = μ_{2} (β - θ) μ_{2} θ + μ_{1} α = (μ_{2} - μ_{1}) β μ_{2} tan θ + μ_{1} tan α = (μ_{2} - μ_{1}) tan β μ_{2} \frac{MN}{N I} + μ_{1} \frac{MN}{NP} = (μ_{2} - μ_{1}) \frac{MN}{NC} \frac{μ _{2}}{v} - \frac{μ _{1}}{u} = \frac{μ _{2} - μ _{1}}{R}$

[For small aperture P & N are the same point]

Note:

(1).It is not always necessary that for a convex boundary the parallel rays always converge. Similarly,for concave boundary the incident parallel ray may converge or diverge depending upon the refractive index of two media,

(2).Laws of refraction are valid for spherical surfaces also.

(3) Pole, centre of curvature, Radius of curvature, Principal axis etc. are defined as spherical mirrors except for the focus.

5.0Key Points Spherical Refraction

Rays should be paraxial.
$μ_{2}$ is the medium in which light rays enter after refraction and $μ_{1}$ is the medium in which light rays move before refraction.
v, u and R(whatever is given) should be put along with a sign.

6.0Focal Length in Spherical Surface

A single spherical surface has two principal focal points which are as follows

First Focus: The first principal focus is the point on the axis where an object should be placed so that the image is formed at infinity.

$u = f_{1}, v = \infty t h e n f ro m \frac{μ _{2}}{v} - \frac{μ _{1}}{u} = \frac{μ _{2} - μ _{1}}{R}$

$- \frac{μ _{1}}{f _{1}} = \frac{μ _{2} - μ _{1}}{R} \Rightarrow f_{1} = \frac{- μ _{1} R}{( μ _{2} - μ _{1} )}$

Second Focus:The second principal focus is the point where parallel rays get focused.

$u = - \infty, v = f_{2}$ then

$\frac{μ _{2}}{f _{2}} = \frac{μ _{2} - μ _{1}}{R} \Rightarrow f_{2} = \frac{μ _{2} R}{( μ _{2} - μ _{1} )}$

Ratio of Focal lengths:

$f_{1} = \frac{- μ _{1} R}{( μ _{2} - μ _{1} )} f_{2} = \frac{μ _{2} R}{( μ _{2} - μ _{1} )} \frac{f _{1}}{f _{2}} = - \frac{μ _{1}}{μ _{2}} or \frac{f _{1}}{μ _{1}} + \frac{f _{2}}{μ _{2}} = 0$

Illustration-1.Find out position and nature of image.

Illustration 1 - Refraction at Spherical Surfaces

Solution:

$μ_{2} = 1.5, μ_{1} = 1, u = - 30 cm, R = + 10 cm \frac{μ _{2}}{v} - \frac{μ _{1}}{u} = \frac{μ _{2} - μ _{1}}{R} \frac{1.5}{v} - \frac{1}{- 30} = \frac{1.5 - 1}{10} \frac{3}{2 v} = \frac{1}{20} - \frac{1}{30} \Rightarrow v = + 90 cm (Real Image)$

Illustration-2.Find out apparent distance of fish as seen by observer. (The radius of the sphere is 10 cm)?

Illustration 2 of Refraction at Spherical Surfaces

Solution:

$μ_{2} = 1, μ_{1} = \frac{4}{3}, u = - 4 cm, R = - 10 cm \frac{μ _{2}}{v} - \frac{μ _{1}}{u} = \frac{μ _{2} - μ _{1}}{R} \frac{1}{v} - \frac{4}{3 ( - 4 )} = \frac{1 - \frac{4}{3}}{- 10} \frac{1}{v} + \frac{1}{3} = \frac{1}{30} \Rightarrow \frac{1}{v} = \frac{1}{30} - \frac{1}{3} v = \frac{- 30}{9} = - 3.34 cm (virtual image)$

Illustration-3.A thin spherical fish bowl with a radius of 10 cm is completely filled with water (refractive index = 4/3). A small fish is located 4 cm away from the center CCC of the bowl. Neglecting the thickness of the glass, determine the apparent position of the fish when viewed from:

(a) Point E (along the left side of the sphere)
(b) Point F (along the right side of the sphere)

Where will the fish appear to be located in each case, as seen by an observer looking through the water–air interface?

Solution:In case of refraction from curved surface

$\frac{μ _{2}}{v} - \frac{μ _{1}}{u} = \frac{μ _{2} - μ _{1}}{R}$

(a)

Where will the fish appear in first case

$μ_{1} = \frac{4}{3}, μ_{2} = 1, R = - 10 cm u = - (10 - 4) = - 6 cm \frac{1}{v} - \frac{( \frac{4}{3} )}{( - 6 )} = \frac{1 - ( \frac{4}{3} )}{( - 10 )} v = - \frac{90}{17} = - 5.3 cm$

i.e. fish will appear at a distance 5.3 cm from E towards

F (lesser than actual distance i.e., 6 cm)

(b)

$μ_{1} = \frac{4}{3}, μ_{2} = 1, R = - 10 cm u = - (10 + 4) = - 14 cm \frac{1}{v} - \frac{( \frac{4}{3} )}{( - 14 )} = \frac{1 - ( \frac{4}{3} )}{( - 14 )} \Rightarrow v = - 16.15 cm$

i.e. fish will appear at a distance 16.15 cm from F towards E,

(more than actual distance, i.e., 14 cm)

Illustration-4.There are two objects O_1 and O_2 at the same distance of 10 cm from a spherical refracting boundary of radius 30 cm as shown in Fig. If the indices of refraction of the media on two sides of the boundary are 1 and (4/3) respectively, (a) where will object $O_{1}$ will appear if seen from $O_{2}$ and (b) where will object $O_{2}$ appear if seen from $O_{1}$ ?

Solution:

(a)

$μ_{1} = 1, μ_{2} = (\frac{4}{3}), R = - 30 cm u_{1} = - 10 cm \frac{( 4/3 )}{v _{1}} - \frac{1}{( - 10 )} = \frac{( 4/3 ) - 1}{( - 30 )} v_{1} = 12 cm$

i.e., object $O_{1}$ (which is 10 cm from P) will appear at 12 cm from P towards C, when seen from $O_{2}$ .

(b)

$μ_{1} = (\frac{4}{3}), μ_{2} = 1, R = 30 cm u_{2} = - 10 cm \frac{1}{v _{2}} - \frac{( 4/3 )}{( - 10 )} = \frac{1 - ( 4/3 )}{( 30 )} v_{2} = - \frac{90}{13} = - 6.92 cm$

i.e., object $O_{2}$ (which is 10 cm from p) will appear at a distance of 6.92 cm. from P

Towards $O_{2}$ . when seen from $O_{1}$ .

Refraction at Spherical Surfaces

Refraction at a spherical surface happens when light passes from one transparent medium to another through a curved, spherical boundary. This bending of light changes the direction and position of the image formed. Unlike refraction at a flat surface, the curvature causes the light rays to converge or diverge, creating real or virtual images.This phenomenon is important in the functioning of lenses and many optical instruments. By studying refraction at spherical surfaces, we can understand how images are formed and learn to calculate the position and size of these images using the laws of optics.

1.0Law of Refraction At Spherical Surface

When light passes from a medium of refractive index $μ_{1}$ to a medium of refractive index 2 by a spherical surface of radius of curvature R then the relation between object distance u and image distance v is given by

$\frac{μ _{2}}{v} - \frac{μ _{1}}{u} = \frac{μ _{2} - μ _{1}}{R}$

2.0Terms Related To Refraction At Spherical Surfaces

(1) Centre of curvature (C)-It is the centre of a sphere of which the surface is a part.

(2) Radius of curvature (R)-It is the radius of the sphere of which the surface is a part.

(3) Pole (P)-It is the geometrical centre of the spherical refracting surface.

(4) Principal Axis-It is the straight line joining the centre of curvature to the pole.

(5) Focus (F)-When a narrow beam of rays of light, parallel to the principal axis and close to it (known as paraxial rays) is incident on a refracting spherical boundary, the refracted beam is found to converge or appear to diverge [depending upon the nature of the boundary (concave/convex) and the refractive index of two media] from a point on the principal axis. This point is called focus.

3.0Cartesian Sign Convention

(1) All distances are measured from the pole (P).

(2) Distances measured in the direction of incident rays are taken as positive.

(3) Distances above the principal axis are taken as positive.

4.0Derivation of Spherical Refraction

Suppose AB is the spherical surface which separates the two medium of refractive index $μ_{1}$ and $μ_{2}$
O and I are the position of object and image in the medium of R.I. $μ_{1}$ and $μ_{2}$

Respectively.

From $△ MOC \Rightarrow i = α + β$

$Δ MC I \Rightarrow β = r + θ$

For small aperture, i and r are small

from Snell’s law,

$μ_{1} sin i = μ_{2} sin r μ_{1} (α + β) = μ_{2} (β - θ) μ_{2} θ + μ_{1} α = (μ_{2} - μ_{1}) β μ_{2} tan θ + μ_{1} tan α = (μ_{2} - μ_{1}) tan β μ_{2} \frac{MN}{N I} + μ_{1} \frac{MN}{NP} = (μ_{2} - μ_{1}) \frac{MN}{NC} \frac{μ _{2}}{v} - \frac{μ _{1}}{u} = \frac{μ _{2} - μ _{1}}{R}$

[For small aperture P & N are the same point]

Note:

(1).It is not always necessary that for a convex boundary the parallel rays always converge. Similarly,for concave boundary the incident parallel ray may converge or diverge depending upon the refractive index of two media,

(2).Laws of refraction are valid for spherical surfaces also.

(3) Pole, centre of curvature, Radius of curvature, Principal axis etc. are defined as spherical mirrors except for the focus.

5.0Key Points Spherical Refraction

Rays should be paraxial.
$μ_{2}$ is the medium in which light rays enter after refraction and $μ_{1}$ is the medium in which light rays move before refraction.
v, u and R(whatever is given) should be put along with a sign.

6.0Focal Length in Spherical Surface

A single spherical surface has two principal focal points which are as follows

First Focus: The first principal focus is the point on the axis where an object should be placed so that the image is formed at infinity.

$u = f_{1}, v = \infty t h e n f ro m \frac{μ _{2}}{v} - \frac{μ _{1}}{u} = \frac{μ _{2} - μ _{1}}{R}$

$- \frac{μ _{1}}{f _{1}} = \frac{μ _{2} - μ _{1}}{R} \Rightarrow f_{1} = \frac{- μ _{1} R}{( μ _{2} - μ _{1} )}$

Second Focus:The second principal focus is the point where parallel rays get focused.

$u = - \infty, v = f_{2}$ then

$\frac{μ _{2}}{f _{2}} = \frac{μ _{2} - μ _{1}}{R} \Rightarrow f_{2} = \frac{μ _{2} R}{( μ _{2} - μ _{1} )}$

Ratio of Focal lengths:

$f_{1} = \frac{- μ _{1} R}{( μ _{2} - μ _{1} )} f_{2} = \frac{μ _{2} R}{( μ _{2} - μ _{1} )} \frac{f _{1}}{f _{2}} = - \frac{μ _{1}}{μ _{2}} or \frac{f _{1}}{μ _{1}} + \frac{f _{2}}{μ _{2}} = 0$

Illustration-1.Find out position and nature of image.

Solution:

$μ_{2} = 1.5, μ_{1} = 1, u = - 30 cm, R = + 10 cm \frac{μ _{2}}{v} - \frac{μ _{1}}{u} = \frac{μ _{2} - μ _{1}}{R} \frac{1.5}{v} - \frac{1}{- 30} = \frac{1.5 - 1}{10} \frac{3}{2 v} = \frac{1}{20} - \frac{1}{30} \Rightarrow v = + 90 cm (Real Image)$

Illustration-2.Find out apparent distance of fish as seen by observer. (The radius of the sphere is 10 cm)?

Solution:

$μ_{2} = 1, μ_{1} = \frac{4}{3}, u = - 4 cm, R = - 10 cm \frac{μ _{2}}{v} - \frac{μ _{1}}{u} = \frac{μ _{2} - μ _{1}}{R} \frac{1}{v} - \frac{4}{3 ( - 4 )} = \frac{1 - \frac{4}{3}}{- 10} \frac{1}{v} + \frac{1}{3} = \frac{1}{30} \Rightarrow \frac{1}{v} = \frac{1}{30} - \frac{1}{3} v = \frac{- 30}{9} = - 3.34 cm (virtual image)$

Illustration-3.A thin spherical fish bowl with a radius of 10 cm is completely filled with water (refractive index = 4/3). A small fish is located 4 cm away from the center CCC of the bowl. Neglecting the thickness of the glass, determine the apparent position of the fish when viewed from:

(a) Point E (along the left side of the sphere)
(b) Point F (along the right side of the sphere)

Where will the fish appear to be located in each case, as seen by an observer looking through the water–air interface?

Solution:In case of refraction from curved surface

$\frac{μ _{2}}{v} - \frac{μ _{1}}{u} = \frac{μ _{2} - μ _{1}}{R}$

(a)

$μ_{1} = \frac{4}{3}, μ_{2} = 1, R = - 10 cm u = - (10 - 4) = - 6 cm \frac{1}{v} - \frac{( \frac{4}{3} )}{( - 6 )} = \frac{1 - ( \frac{4}{3} )}{( - 10 )} v = - \frac{90}{17} = - 5.3 cm$

i.e. fish will appear at a distance 5.3 cm from E towards

F (lesser than actual distance i.e., 6 cm)

(b)

$μ_{1} = \frac{4}{3}, μ_{2} = 1, R = - 10 cm u = - (10 + 4) = - 14 cm \frac{1}{v} - \frac{( \frac{4}{3} )}{( - 14 )} = \frac{1 - ( \frac{4}{3} )}{( - 14 )} \Rightarrow v = - 16.15 cm$

i.e. fish will appear at a distance 16.15 cm from F towards E,

(more than actual distance, i.e., 14 cm)

Illustration-4.There are two objects O_1 and O_2 at the same distance of 10 cm from a spherical refracting boundary of radius 30 cm as shown in Fig. If the indices of refraction of the media on two sides of the boundary are 1 and (4/3) respectively, (a) where will object $O_{1}$ will appear if seen from $O_{2}$ and (b) where will object $O_{2}$ appear if seen from $O_{1}$ ?

Solution:

(a)

$μ_{1} = 1, μ_{2} = (\frac{4}{3}), R = - 30 cm u_{1} = - 10 cm \frac{( 4/3 )}{v _{1}} - \frac{1}{( - 10 )} = \frac{( 4/3 ) - 1}{( - 30 )} v_{1} = 12 cm$

i.e., object $O_{1}$ (which is 10 cm from P) will appear at 12 cm from P towards C, when seen from $O_{2}$ .

(b)

$μ_{1} = (\frac{4}{3}), μ_{2} = 1, R = 30 cm u_{2} = - 10 cm \frac{1}{v _{2}} - \frac{( 4/3 )}{( - 10 )} = \frac{1 - ( 4/3 )}{( 30 )} v_{2} = - \frac{90}{13} = - 6.92 cm$

i.e., object $O_{2}$ (which is 10 cm from p) will appear at a distance of 6.92 cm. from P

Towards $O_{2}$ . when seen from $O_{1}$ .

Frequently Asked Questions

Why does refraction occur at a spherical surface, and how is it different from refraction at a plane surface?

Why is it that in spherical refraction, magnification may not be uniform in all directions for large objects?

How does spherical aberration impact the accuracy of the refraction formula?

Can a spherical refracting surface act as both converging and diverging under different conditions?

Why is the concept of “optical center” not applicable to single spherical refracting surfaces?

Join ALLEN!

Refraction at Spherical Surfaces

1.0Law of Refraction At Spherical Surface

2.0Terms Related To Refraction At Spherical Surfaces

3.0Cartesian Sign Convention

4.0Derivation of Spherical Refraction

5.0Key Points Spherical Refraction

6.0Focal Length in Spherical Surface

Table of Contents

Frequently Asked Questions

Why does refraction occur at a spherical surface, and how is it different from refraction at a plane surface?

Why is it that in spherical refraction, magnification may not be uniform in all directions for large objects?

How does spherical aberration impact the accuracy of the refraction formula?

Can a spherical refracting surface act as both converging and diverging under different conditions?

Why is the concept of “optical center” not applicable to single spherical refracting surfaces?

Join ALLEN!

Refraction at Spherical Surfaces

1.0Law of Refraction At Spherical Surface

2.0Terms Related To Refraction At Spherical Surfaces

3.0Cartesian Sign Convention

4.0Derivation of Spherical Refraction

5.0Key Points Spherical Refraction

6.0Focal Length in Spherical Surface

Table of Contents