Refraction at Spherical Surfaces is the fundamental concept that helps us understand the design and working of lenses. Before understanding refraction at spherical surfaces, let us know the lenses used. Consider the below diagram representing the refraction of light from a spherical (concave) surface in which the ray of light from the object \(O\) gets refracted and forms a virtual image at \(I.\)

Here, the change in the path of the ray of light from the object \(O\) depends on the shape of the boundary separating the media. This article will understand the definition of refraction of light at spherical surfaces lenses, types of lenses and learn how to derive an expression for refraction at spherical surfaces.

What is Refraction at Spherical Surfaces?

Refraction at spherical surfaces can be well understood when we individually understand each term used in the concept. So, let us first understand the concept of ‘Refraction’ and then get more information about the term ‘Spherical Surfaces’.

Refraction is the phenomenon of change in the path of light while going from one medium to another. This change in path occurs at the boundary of two media. Refraction is caused due to change in the speed of light while going from one medium to other.

Due to refraction, many such phenomena occur in nature, like the twinkling of stars, advanced sunrise, delayed sunset, etc. The fact that lenses can converge or diverge rays of light passing through them is due to the phenomenon of refraction. Due to refraction, we see a pencil broken when dipped in a beaker filled with water. If we look around, we can spot many such occurrences due to refraction.

Spherical surfaces are the surfaces that are part of a sphere. Spherical mirrors are examples of spherical surfaces that reflect the light falling on them. Here, we are going to focus on refraction at spherical surfaces. There are two such spherical surfaces: convex and concave.

A convex surface is a surface that is curved outwards, as shown in the below diagram:

And a concave surface is a surface that is turned inwards, as shown in the below diagram:

Sign Conventions for Refraction at Spherical Surfaces

While studying refraction at spherical surfaces, we follow the below-mentioned sign convention:

1. All distances are measured from the pole of the spherical refracting surface.
2. The distances measured in the direction of incidence of light are taken as positive, and the distances measured in a direction opposite to the direction of incidence of light are taken as negative.
3. The distances measured in the perpendicular direction above the principal axis are positive. And the distances measured in the perpendicular direction below the principal axis are negative.

These points can be summarised in the below diagram:

Here, we need to note that when the object faces a convex refracting surface, the radius of curvature \(R\) of the surface is positive. And when the object faces a concave refracting surface, the radius of curvature \(R\) of the surface is negative.

Derivation of Refraction at Spherical Surfaces for Point Objects

We can derive an expression for refraction at spherical surfaces occurs in two ways. The ray of light may travel from a rarer medium to a denser medium in which a ray of light bends towards the normal, or the ray of light may travel from a denser medium to a rarer medium in which the ray of light bends away from normal.

Here, while considering the refraction at spherical surfaces, we assume:

1. The object taken here is point sized and is lying on the principal axis of the spherical refracting surface.
2. The aperture of the spherical refracting surface is small.
3. The incident and the refracted rays make small angles with the principal axis of the spherical surface so that \(\sin i \approx i\) and \(\sin r \approx r.\)

Here, while considering refraction from rarer to denser medium, two cases may occur: refraction from rarer to denser medium at a convex spherical surface and at a concave spherical surface. Similarly, while considering refraction from denser to rarer medium, two cases may occur: refraction from denser to rarer medium at a convex spherical surface and a concave spherical surface. Among all four cases, we may consider any of them to derive the relation governing the refraction at spherical surfaces.

We will take the case of refraction from rarer to denser medium at a convex spherical surface to derive the relation. The ray diagram of such a case is shown below:

Here, let us consider the case of refraction when a real image is formed. In the above diagram, light from the point object \(O\) to another medium with refractive index \({n_1}{n_2}.\) As \({n_1} < {n_2},{n_1}\) is the rarer medium and \({n_2}\) is the denser medium. The boundary of the to another medium with refractive index second medium is convex towards the rarer medium. Let \(P\) be the pole and \(C\) be the centre of the curvature of the refracting spherical surface. So, the radius of curvature of the surface is \(PC = R.\)

The point object \(O\) is lying on the principal axis of the spherical refracting surface. A ray of light passing along the principal axis will pass straight, but a ray of light incident on the spherical refracting surface at \(\angle i\) is refracted at \(\angle r,\) bending towards normal. The two refracted rays meet at \(I,\), where the image is formed. This is the real image of the object \(O.\)

Let the angle formed between the oblique incident ray and the principal axis be \(\alpha, \), the angle formed between the oblique refracted ray and the principal axis be \(\beta,\) and the angle formed between the normal at the point of incidence \(\left( A \right)\) and the principal axis be \(\gamma.\)

Now, let us drop a perpendicular \(\left( {AM} \right)\) from the point of incidence \(\left( A \right)\) to a point \(\left( M \right)\) on the principal axis.

From the ray diagram we get, \(\angle {\rm{AOM}} = \alpha ,{\mkern 1mu} \angle {\rm{AIM}} = \beta\) and \(\angle {\rm{ACM}} = \gamma.\)

As the external angle of a triangle is equal to the sum of the internal opposite angles, so \(\gamma \) is the external angle of the \(\Delta {\rm{ACI}}\) with \(r\) and \(\beta\) as the internal opposite angles.

\(\therefore \,\gamma = r + \beta\) or

\(r = \gamma – \beta\)

Similarly, \(i\) will be the external angle of the \(\Delta {\rm{AOC}}\) with \(\alpha \) and \(\gamma\) as the internal opposite angles.

\(\therefore \,i = \alpha + \gamma\)

According to Snell’s law,

\(\frac{{{n_2}}}{{{n_1}}} = \frac{{\sin i}}{{\sin r}} = \frac{i}{r}\)

Since the angles are small. Substituting the values of \(i\) and \(r,\) we get

\(\frac{{{n_2}}}{{{n_1}}} = \frac{{\alpha + \gamma }}{{\gamma – \beta }}\)

\(\therefore \,{n_2}\left( {\gamma – \beta } \right) = {n_1}\left( {\alpha + \gamma } \right)\)

As the angle, \(\alpha ,\beta \) and \(\gamma\) are small, the aperture of the spherical refracting surface is small, and the point \(\left( M \right)\) of the perpendicular dropped from the point of incidence to the principal axis is close to the pole \(\left( P \right),\) using, \(\theta = \frac{l}{r},\) we get

\(\alpha = \frac{{{\rm{AM}}}}{{{\rm{OM}}}}\)

\(\beta = \frac{{{\rm{AM}}}}{{{\rm{MI}}}}\)

\(\gamma = \frac{{{\rm{AM}}}}{{{\rm{MC}}}}\)

\(\therefore \,{n_2}\left( {\frac{{{\rm{AM}}}}{{{\rm{MC}}}} – \frac{{{\rm{AM}}}}{{{\rm{MI}}}}} \right) = {n_1}\left( {\frac{{{\rm{AM}}}}{{{\rm{OM}}}} + \frac{{{\rm{AM}}}}{{{\rm{MC}}}}} \right)\)

\({n_2}\left( {\frac{1}{{{\rm{MC}}}} – \frac{1}{{{\rm{MI}}}}} \right) = {n_1}\left( {\frac{1}{{{\rm{OM}}}} + \frac{1}{{{\rm{MC}}}}} \right)\)

Now, as \(M\) is close to \(P,\) we Simplifying the equation, we get
get \({\rm{MC}} \approx {\rm{PC}},\,{\rm{MI}} \approx {\rm{PI}}\) and \({\rm{OM}} \approx {\rm{OP}}\)

\(\therefore {n_2}\left( {\frac{1}{{{\rm{PC}}}} – \frac{1}{{{\rm{PI}}}}} \right) = {n_1}\left( {\frac{1}{{{\rm{OP}}}} + \frac{1}{{{\rm{PC}}}}} \right)\)

Using the cartesian sign conventions, we get, \(OP = – u,\,PI = + v\) and the \(PC = + R.\) Putting these values in the above equation, we get
\(\frac{{{n_2}}}{R} – \frac{{{n_2}}}{v} = \frac{{{n_1}}}{{ – u}} + \frac{{{n_1}}}{R}\)

\(\therefore \frac{{{n_1}}}{{ – u}} + \frac{{{n_2}}}{v} = \frac{{{n_2} – {n_1}}}{R}\)

\(\therefore \frac{{{n_2}}}{v} – \frac{{{n_1}}}{u} = \frac{{{n_2} – {n_1}}}{R}\)

This is the relation governing refraction from rarer to denser medium at a convex spherical refracting surface. It can be easily shown that in the case of the refraction from rarer to denser medium at a concave spherical surface, the same relation is obtained. Infact for every possible situation, the same relation is obtained.

Magnification Produced by Refraction at Spherical Surfaces for Extended Objects

Consider the case of light rays coming from an extended object \(AB\) that are getting refracted from a convex refracting surface, as shown in the below ray diagram:

Here, the object \(AB\) is placed perpendicular to the principal axis of the convex spherical surface \(XY.\) The ray originating from \(A\) and going towards \(C\) is incident normally on the spherical surface \(XY,\), so it goes undeviated in the second medium. The image of A must be on the line \(AC.\) The image of \(B\) must be formed on the line \(BPC\) at \(B’.\) If we drop a perpendicular from \(B’\) on \(BPC,\) it will intersect the line \(AC\) at \(A’,\) which will be the image of \(A.\)

Now the lateral magnification for extended objects is given by the relation,
\(m = \frac{{{h_i}}}{{{h_o}}}\)
where
\(m\) is the magnification
\({{h_i}}\) is the image height
\({{h_o}}\) is the object height
From the above ray diagram and using the sign conventions we get,
\(AB = + {h_o}\)
\(A’B’ = – {h_i}\)
Putting these values in the relation of magnification we get,
\(m = \frac{{{h_i}}}{{{h_o}}} = \frac{{ – A’B’}}{{AB}}\)
Now \(\Delta ABC\) and \(\Delta A’B’C\) are similar triangles. So we get,
\(m = – \frac{{A’B’}}{{AB}} = – \frac{{B’C}}{{BC}} = – \frac{{PB’ – PC}}{{PB + PC}}\)
From the above ray diagram and using the sign conventions we get,
\(PB = \, – u\) that is the distance of the object \(AB\) from the pole \(\left( P \right)\) of the spherical surface
\(PC = + R\) that is the radius of curvature of the spherical surface
\(PB’ = + v\) that is the distance of the image \(A’B’\) from the pole \(\left( P \right)\)
of the spherical surface
Putting these values in the above equation, we get
\(m = \, – \frac{{v – R}}{{ – u + R}} = \frac{{R – v}}{{R – u}}\)
We also know that the relation of refraction at spherical surfaces is given by,
\(\frac{{{n_2}}}{v} – \frac{{{n_1}}}{u} = \frac{{{n_2} – {n_1}}}{R}\)
Simplifying it we get,
\(\frac{{{n_2}u – {n_1}v}}{{uv}} = \frac{{{n_2} – {n_1}}}{R}\)
\(\therefore \,R = \frac{{uv\left( {{n_2} – {n_1}} \right)}}{{{n_2}u – {n_1}v}}\)
\(\therefore \,R = \frac{{uv\left( {{n_2} – {n_1}} \right)}}{{{n_2}u – {n_1}v}}\)
This gives,
\(R – v = \frac{{uv\left( {{n_2} – {n_1}} \right)}}{{{n_2}u – {n_1}v}} – v = \frac{{uv\left( {{n_2} – {n_1}} \right) – v\left( {{n_2}u – {n_1}v} \right)}}{{{n_2}u – {n_1}v}}\)
\(\therefore \,R – v = \frac{{{n_2}uv – {n_1}uv – {n_2}uv + {n_1}{v^2}}}{{{n_2}u – {n_1}v}} = \frac{{{n_1}v\left( {v – u} \right)}}{{{n_2}u – {n_1}v}}\)
Similarly,
\(R – u = \frac{{uv\left( {{n_2} – {n_1}} \right)}}{{{n_2}u – {n_1}v}} – u = \frac{{uv\left( {{n_2} – {n_1}} \right) – u\left( {{n_2}u – {n_1}v} \right)}}{{{n_2}u – {n_1}v}}\)
\(\therefore \,R – u = \frac{{{n_2}uv – {n_1}uv – {n_2}{u^2} + {n_1}uv}}{{{n_2}u – {n_1}v}} = \frac{{{n_2}u\left( {v – u} \right)}}{{{n_2}u – {n_1}v}}\)
Now putting the values of \(R – v\) and \(R-u\) in the magnification relation we get,

\(m = \frac{{\frac{{{n_1}v\left( {v – u} \right)}}{{{n_2}u – {n_1}v}}}}{{\frac{{{n_2}u\left( {v – u} \right)}}{{{n_2}u – {n_1}v}}}} = \frac{{{n_1}v\left( {v – u} \right)}}{{{n_2}u – {n_1}v}} \times \frac{{{n_2}u – {n_1}v}}{{{n_2}u\left( {v – u} \right)}} = \frac{{{n_1}v}}{{{n_2}u}}\)

Thus, the relation of magnification produced by refraction at spherical surfaces for extended objects is given by,
\(m = \frac{{{n_1}v}}{{{n_2}u}}\)
This relation holds good for any single refracting surface, convex or concave.

Refraction at Spherical Surfaces Solved Examples

Example 1
Light from a point source in the air falls on a convex spherical glass surface with \(n = 1.5\) and \(R = 40\,{\rm{cm}}.\) Calculate the position of the image when the light source is at \(1.2\,{\rm{m}}\) from the glass surface.
Solution:
Given that,
The refractive index of air is \({n_1} = 1\)
The refractive index of glass is \({n_2} = 1.5\)
The radius of curvature of the convex spherical surface is \(R = + \,40{\mkern 1mu} {\rm{cm}}\)
The object distance is \(u = – 1.2\,{\rm{m}} = – 120\,{\rm{cm}}\)
The image distance is given by the relation,
\(\frac{{{n_1}}}{{ – u}} + \frac{{{n_2}}}{v} = \frac{{{n_2} – {n_1}}}{R}\)
\(\therefore \frac{1}{{ – \left( { – 120} \right)}} + \frac{{1.5}}{v} = \frac{{1.5 – 1}}{{40}}\)
\(\frac{1}{{120}} + \frac{{1.5}}{v} = \frac{{0.5}}{{40}}\)
\(\frac{{1.5}}{v} = \frac{{0.5}}{{40}} – \frac{1}{{120}} = \frac{{1.5 – 1}}{{120}} = \frac{{0.5}}{{120}}\)
\(\therefore v = \frac{{1.5 \times 120}}{{0.5}} = 360\,{\rm{cm}} = 3.6\,{\rm{m}}\)
Thus, the image is formed at \(3.6\,{\rm{m}}\) from the pole of the convex refracting glass surface in the direction of incidence of light.

Example 2
Find the radius of curvature of a concave refracting surface of refractive index \(n = \frac{3}{2}\) that can form a virtual image at \(20\,{\rm{cm}}\) of an object kept at a distance of \(40\,{\rm{cm}}\) in the same medium. The light from the medium is getting refracted into the air.
Solution:
Given that,
The refractive index of the medium is ({n_2} = frac{3}{2} = 1.5)
The refractive index of air is ({n_1} = 1)
The object distance is (u = , – 40,{rm{cm}})
The image distance is (v = , – 20,{rm{cm}})
The radius of curvature is given by the relation,
(frac{{{n_2}}}{{ – u}} + frac{{{n_1}}}{v} = frac{{{n_1} – {n_2}}}{R})
(therefore frac{{1.5}}{{ – left( { – 40} right)}} + frac{1}{{ – 20}} = frac{{1 – 1.5}}{R})
(frac{{1.5}}{{40}} – frac{1}{{20}} = frac{{ – 0.5}}{R})
(therefore frac{{1.5 – 2}}{{40}} = frac{{ – 0.5}}{R})
(frac{{ – 0.5}}{{40}} = frac{{ – 0.5}}{R})
(therefore,R = 40,{rm{cm})
Thus, the radius of curvature of the concave refracting surface is (40,{rm{cm}})

Applications of Refraction at Spherical Surfaces

Refraction at spherical surfaces finds application in many situations. Some of them are as under:
1. It helps us to estimate the behaviour of the rays of light passing through various lenses like those given in the below picture:

2. It helps us understand how light rays will behave while entering the second medium with varying refractive index.
3. Combining the relations of refraction at spherical surfaces of both the lens surfaces, we get the formula of a lens as a whole entity.

FAQs on Refraction at Spherical Surface

Q.1. What is the magnification equation for refraction at spherical surfaces?
Ans: The magnification equation for refraction at spherical surfaces is \(m = \frac{{{h_i}}}{{{h_o}}} = \frac{{{n_1}v}}{{{n_2}u}}\)

Q.2. What is refraction?
Ans: Refraction is the phenomenon of bending of the ray of light at the interface of two media while the light enters the second medium with different optical densities.

Q.3. What are spherical surfaces?
Ans: Spherical surfaces are the surfaces that are part of a sphere.

Q.4. What is the relation of refraction at spherical surfaces when the object lies in the rarer medium?
Ans: The relation governing refraction at spherical surfaces when the object lies in the rarer medium is (frac{{{n_1}}}{{ – u}} + frac{{{n_2}}}{v} = frac{{{n_2} – {n_1}}}{R})

Q.5. What is the relation of refraction at spherical surfaces when the object lies in the denser medium?
Ans: The relation governing refraction at spherical surfaces when the object lies in the denser medium is (frac{{{n_2}}}{{ – u}} + frac{{{n_1}}}{v} = frac{{{n_1} – {n_2}}}{R})