Refraction At Spherical Surface
Refraction at a spherical surface is a fundamental concept in optics, where light rays change direction as they pass through the boundary between two different media with varying refractive indices. Unlike plane surfaces, spherical surfaces can converge or diverge light rays, leading to the formation of real or virtual images. This phenomenon is governed by Snell's law and the geometry of the spherical surface, making it crucial for understanding the behaviour of lenses and optical instruments. In real life, refraction at spherical surfaces is observed in devices like eyeglasses, cameras, microscopes, and telescopes, where precise control over light paths is essential for clear vision and image formation. The principles of this refraction also play a role in natural phenomena, such as the focusing effect of water droplets on a surface or the formation of mirages. In this article, we will cover the concept of Refraction at a spherical surface it is a fundamental concept, that impacts how light rays bend when transitioning between media with different refractive indices, some important points for solving numerical questions and some solved examples for better clarity is given in this article, read on to know in detailed.
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- Refraction at Spherical Surface
- Lateral Magnification For Refracting Spherical Surface
- Solved Example Based on Refraction at Spherical Surface
- Summary
Refraction at Spherical Surface
Refraction at a spherical surface is a key concept that describes how light rays bend when passing through a curved boundary between two media with different refractive indices. Unlike refraction at flat surfaces, spherical surfaces can either converge or diverge light rays due to their curvature. This bending of light is described by Snell's law and the radius of curvature of the spherical surface
If an object 0 is placed in front of a curved surface as shown in the above figure, then the Refraction formula is given as
$\frac{n_2}{v}-\frac{n_1}{u}=\frac{n_2-n_1}{R}$
where
$n_1=$ Refractive index of the medium from which light rays are coming (from the object).
$n_2=$ Refractive index of the medium in which light rays are entering.
and $n_1<n_2$
and u = Distance of object, v = Distance of image, R = Radius of curvature
Note:
- use sign convention while solving the problem
- Real image forms on the side of a refracting surface that is opposite to the object,
and virtual image forms on the same side as the object. - Using $R=\infty$ (i.e for plane surface)
we get $\frac{n_2}{v}=\frac{n_1}{u} \Rightarrow \frac{n_2}{n_1}=\frac{v}{u}$
Lateral Magnification For Refracting Spherical Surface
Lateral magnification for a refracting spherical surface refers to the ratio of the height of the image to the height of the object. For a refracting spherical surface, this magnification can be determined using the distances of the object and the image from the refracting surface.
If an object AB is placed in front of a curved surface as shown in the above figure, then the lateral Magnification formula is given as
Lateral magnification, $m=\frac{\text { Image height }}{\text { Object height }}=\frac{-\left(A^{\prime} B^{\prime}\right)}{A B}$
$
m=-\frac{A^{\prime} B^{\prime}}{A B}=-\frac{\mu_1}{\mu_2} \times \frac{v}{u}=-\frac{v / \mu_2}{u / \mu_1}
$
where,
$\mu_{1=}$=Refractive index of the medium from which light rays are coming (from the object).
$\mu_{2=}$=Refractive index of the medium in which light rays are entering.
and $\mu_1<\mu_2$
and $\mathrm{u}=$ Distance of object, $\mathrm{v}=$ Distance of image, $\mathrm{R}=$ Radius of curvature
Solved Example Based on Refraction at Spherical Surface
Example 1: A small object is embedded in a glass sphere $(\mu=1.5)$ of radius 5cm at a distance 1.5cm left of the centre. The image of the object as seen by an observer standing to the left of the sphere is at a distance:
1) 3cm from center
2) 2.5cm from center
3)1.5cm from center
4) 3.5cm from center
Solution:
Refraction through a spherical surface
$\frac{\mu_2}{v}-\frac{\mu_1}{u}=\frac{\mu_2-\mu_1}{R}$
wherein
$u=$ Object distance
$v=$ image distance
$R=$ Radius of curvature
Hence, the image is formed 3cm to the right of the A or 2cm to the left of the C.
Hence, the answer is the option (1).
Example 2: The eye can be regarded as a single refracting surface. The radius of curvature of this surface is equal to that of the cornea (7.8mm). This surface separates two media of refractive indices 1 and 1.34. Calculate the distance from the refracting surface at which a parallel beam of light will come to focus.
1) 4.0 cm
2) 2 cm
3) 3.1 cm
4) 1.0 cm
Solution:
Refraction through a spherical surface
Refraction through a spherical surface
$
\begin{aligned}
& \frac{\mu_2}{v}-\frac{\mu_1}{u}=\frac{\mu_2-\mu_1}{R} \\
& \left(\frac{1.34}{V}-\frac{1}{\infty}\right)=\frac{1.34-1}{7.8} \\
& V=30.7 \mathrm{~mm} \approx 3.1 \mathrm{~cm}
\end{aligned}
$
Hence, the answer is the option (3).
Example 3: The figure shows a solid glass sphere of radius 5 cm that has a small air bubble O trapped at a distance 2 cm from the centre C . The refractive index of the material of glass is 1.5 .Find the apparent position (in cm) of the bubble from point P, when seen through the surface of the sphere from an outside point P,
1) 5
2) 2.5
3) 7.5
4) 10
Solution:
Here rays are striking on concave side of the sphere
Using, $\frac{\mu_2}{v}-\frac{\mu_1}{u}=\frac{\left(\mu_2-\mu_1\right)}{R}$
Here, $u=-\stackrel{u}{P} O=-(P C-O C)=-(5-2) \mathrm{cm}=-3 \mathrm{~cm}$
$R=-5 \mathrm{~cm} ;$ and $\mu_1=\frac{3}{2}$ and $\mu_2=1$
$
\frac{1}{v}-\frac{3 / 2}{(-3)}=\frac{(1-3 / 2)}{(-5)} \Rightarrow v=-\frac{5}{2} \mathrm{~cm}=-2.5 \mathrm{~cm}
$
Hence, the answer is the option (2).
Example 4: A parallel beam of light is allowed to fall on a transparent spherical globe of diameter 30 cm and refractive index 1.5. The distance from the centre of the globe at which the beam of light can converge is ___________
1) 225
2) 114
3) 324
4) 12
Solution:
$1^{\text {st }}$ refraction
$
\begin{aligned}
& \frac{1.5}{\mathrm{v}}-\frac{1}{\infty}=\frac{1.5-1}{+15} \\
& \mathrm{~V}_1=45 \mathrm{~cm}
\end{aligned}
$
For $2^{\text {nd }}$ refraction
$
\begin{aligned}
& \frac{1}{V_2}-\frac{1.5}{+15}=\frac{1-1.5}{-15} \\
& \frac{1}{V_2}=\frac{1}{10}+\frac{1}{30}=\frac{4}{30} \\
& V_2=+7.5 \mathrm{~cm}
\end{aligned}
$
Distance from centre
$
\begin{aligned}
& =15+7.5 \\
& =22.5 \mathrm{~cm}=225 \mathrm{~mm}
\end{aligned}
$
Hence, the answer is the option (1).
Example 5: The electrostatic force between the metal plates of an isolated parallel plate capacitor C having a charge Q and area A, is
1) proportional to the square root of the distance between the plates.
2) linearly proportional to the distance between the plates.
3) Independent of the distance between the plates.
4) Inversely proportional to the distance between the plates.
Solution:
As we have learned
The force between Parallel Plates' Capacitor
$
F=\frac{\sigma^2 A}{2 \epsilon_0}=\frac{Q^2}{2 \epsilon_0 A}=\frac{C V^2}{2 d}
$
wherein
$\sigma-$ Surface charge density.
$
F=\frac{\sigma^2 A}{2 \epsilon_0}
$
F is independent of the distance between the plates.
Hence, the answer is the option (3).
Summary
Refraction at spherical surfaces is a critical concept in optics that explains how light rays bend at curved boundaries between different media, forming real or virtual images. Governed by Snell's law and the surface's curvature, this principle is essential for understanding the functioning of lenses and various optical instruments. Lateral magnification for such surfaces is the ratio of image height to object height and can be calculated using the refractive indices and distances involved. Solved examples demonstrate these principles in practical scenarios, reinforcing their application and understanding.