26.4 REFRACTION AT SPHERICAL SURFACE
26.4.1 Formula for Refraction at Spherical Surface
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Figure 26.20 A point object O is placed in a medium with refractive index n₁ near the spherical surface SS'. When light rays from O reach the surface, they refract into the second medium with refractive index n₂ (n₂ > n₁). If the refracted rays converge, a real image I is formed in the second medium. However, if the rays diverge after refraction, a virtual image I is formed, which appears to originate from the first medium as shown by the extension of the refracted rays. |
1. Suppose a spherical surface SS' separates one transparent medium of refractive index n1 from another transparent medium of refractive index n2 where n2 > n1. A point object O is placed in the medium of refractive index n1 close to the surface, as shown in Figure 26.20(a).
2. When light from O reaches the surface, it will refract at the curved surface as it enters the second medium. After that, if the light is allowed to converge to a point, a real point image I will be formed.
3. On the other hand, if the light diverges after entering the second medium, a virtual point image will be formed, as shown in Figure 26.20(b), (c).
4. The spherical surface has radius of curvature r. The point object is placed at distance u from the surface. An image is formed at distance v from the surface. Then u, v and r are related to one another by the following formula:
where |n2 − n1| is a positive value.
Derivation:
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| Figure 26.21 The diagram shows the geometry used to derive the formula for refraction at a spherical surface. A point object O is located at a distance u from the surface, while the image I is formed at a distance v. The spherical surface has a radius of curvature r with center C. Angles i and r represent the angles of incidence and refraction, respectively, while α, β, and γ are small angles used in the approximation for the derivation of the refraction formula. |
Refer to Figure 26.21.
Using ΔOAC,
Using ΔCAI,
i = α + γ ...(1)
γ = r + β
r = γ − β ...(2)
At A,
n1 sin i = n2 sin r ...(3)
If A is close to the axis of the curved surface then i and r are very small. We can assume
sin i ≅ i
and
sin r ≅ r
where i and r are in radian.
Expression (3) becomes
n1 i = n2 r
Using (1) and (2),
n1(α + γ) = n2(γ − β)
Rearranging,
n1α + n2β = (n2 − n1)γ ...(4)
tan α ≅ α = AQ/OQ, tan β ≅ β = AQ/IQ, tan γ ≅ γ = AQ/CQ
since i and r are small, and so α, β and γ are also small.
From (4),
(n1AQ / OQ) + (n2AQ / IQ) = (n2 − n1)AQ / CQ
But, OQ ≅ OP = u, IQ ≅ IP = v, CQ ≅ CP = r
We get
n1/u + n2/v = (n2 − n1)/r
26.4.2 Sign Convention
(a) For u and v
We adopt the sign convention of ‘real is positive’.
If the object and image are real, then the object distance u and image distance v are positive.
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| The diagram illustrates the sign convention for the radius of curvature r at a spherical surface. In Figure (a), the centre of curvature C lies in the medium with refractive index n₂ where n₂ > n₁, so r is taken as positive. (b), the centre of curvature C lies in the medium with refractive index n₂ where n₂ < n₁, so r is taken as negative. |
(b) For r
Refer to Figure 26.22(a). The centre of curvature C of the surface is inside the medium whose refractive index is n2 where n2 > n1. Then r is positive.
Refer to Figure 26.22(b). C is inside the medium with n2 where n2 < n1. Then r is negative.
EXAMPLE 26.8
Refer to Figure 26.22(a), n1 = 1.0 and n2 = 1.52. The radius of curvature of the spherical surface is 5.0 cm. A point object is placed (a) 50 cm (b) 5 cm from the surface and inside the medium with refractive index n1. Determine the distance of the image formed.
Answer
(a)
n1/u + n2/v = (n2 − n1)/r
1/(+50) + 1.52/v = (1.52 − 1.0)/(+5.0)
v = +18.1 cm (real image)
The real image is formed at a distance of 18.1 cm from the spherical surface and inside the medium with n2. The object and image are on opposite sides of the surface.
(b)
1/(+5) + 1.52/v = (1.52 − 1.0)/(+5.0)
v = −15.8 cm (virtual image)
The virtual image is formed at a distance of 15.8 cm from the spherical surface and inside the medium with n1. The object and image are in the same medium.
EXAMPLE 26.9
Refer to Figure 26.23. A point object O is placed 10 cm in front of a spherical surface SS' with radius of curvature 10 cm. Determine the image distance.
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Figure 26.23 A point object O is placed 10 cm in front of a spherical surface SS' with a radius of curvature of 10 cm. The diagram shows the refraction of light at the curved surface and the formation of the image I, used to determine the image distance based on the refraction formula. |
Answer
Since the object is in an optically denser medium (n1 > n2), u = +10 cm and C is inside an optically less dense median, we have r = −10 cm.
n1/u + n2/v = |n2 − n1| / r
1.5/(+10) + 1.0/v = |1.0 − 1.5|/(−10)
v = −5.0 cm (virtual image)
The virtual image is formed at a distance of 5.0 cm from the spherical surface and inside the medium with n1. The object and image are in the same medium.
Applications of Refraction at a Spherical Surface
Understanding how curved surfaces bend light and form images in real-life technology
🔍 Introduction
Refraction at a spherical surface is an important concept in optics. It explains how light bends when passing between two media with different refractive indices through a curved boundary. This principle is widely applied in many optical devices and everyday situations.
👓 Lenses (Convex & Concave)
Lenses are made from spherical surfaces that refract light to form images.
- Convex lenses focus light to form real images (used in cameras, microscopes).
- Concave lenses spread light (used in glasses for myopia).
📷 Optical Instruments
Many devices rely on spherical refraction to function properly:
- Microscopes magnify tiny objects
- Telescopes observe distant objects
- Cameras capture focused images
👁️ Human Eye
The human eye uses spherical refraction naturally.
- The cornea and lens bend light
- Images are formed on the retina
- Vision defects occur due to improper refraction
💧 Water & Glass Effects
Everyday examples of spherical refraction include:
- Objects in water appear bent or shifted
- Fish bowls distort or magnify images
🔌 Optical Fiber
Refraction occurs when light enters and exits optical fibers, enabling data transmission in communication systems.
✨ Conclusion: Refraction at a Spherical Surface
Refraction at a spherical surface is a fundamental concept in optics that explains how light bends when passing between different media through a curved boundary. This principle is essential for understanding image formation in lenses, optical instruments, and even the human eye.
By applying the formula n₁/u + n₂/v = (n₂ − n₁)/r, we can accurately determine the position and nature of images formed by spherical surfaces. This makes it a key concept in physics education, optical engineering, and real-world technology.
From cameras and microscopes to vision correction and fiber optics, refraction at spherical surfaces plays a crucial role in modern science and daily life. Mastering this concept not only improves problem-solving skills but also provides deeper insight into how light behaves in practical applications.
❓ Frequently Asked Questions (FAQ)
Refraction at a spherical surface is the bending of light when it passes between two media with different refractive indices through a curved boundary. This process determines how images are formed.
The formula is: n₁/u + n₂/v = (n₂ − n₁)/r, where u is object distance, v is image distance, and r is the radius of curvature.
A real image is formed when light rays actually converge, while a virtual image is formed when rays appear to diverge from a point.
The sign convention helps determine whether distances and radius are positive or negative. Typically, real objects/images are positive, and the sign of r depends on the position of the center of curvature.
It is used in lenses, cameras, microscopes, telescopes, eyeglasses, and the human eye to form and control images.
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