Pole, P
The geometric centre of the refracting surface. All object and image distances are measured from P.
Refraction at Spherical Surface
Concepts, Cartesian sign convention, paraxial derivation, five ray-diagram cases, solved numericals, international exam practice and rapid revision.
Class 12 Physics
CBSE
NEET
JEE
IB · A-Level · AP
1
Refraction Through a Spherical Refracting Surface
A spherical refractor is a curved boundary separating two transparent media of refractive indices μ1 and μ2. The boundary is a part of a sphere.
Centre, C
The centre of the parent sphere. The line PC is the principal axis and PC = R.
Paraxial Rays
Rays close to the principal axis and making small angles with it. The standard formula is valid in this approximation.
Standard spherical refraction formula
μ2/v − μ1/u = (μ2 − μ1)/R
Here, u is object distance, v is image distance, and R is radius of curvature. Every value must carry its Cartesian sign.
2
Convex and Concave Spherical Surfaces
Convex Spherical Surface
As seen by the incident light, the surface bulges towards the object. If light travels left to right and C lies to the right of P, then R > 0.
Rarer → denser: usually converging. A distant object can form a real image in the denser medium.
Concave Spherical Surface
As seen by the incident light, the surface caves away from the object. If light travels left to right and C lies to the left of P, then R < 0.
Rarer → denser: usually diverging. The refracted rays may appear to arise from a virtual image.
Rarer to Denser
μ2 > μ1. The ray bends towards the normal. Optical power depends on both the index difference and the sign of R.
Denser to Rarer
μ2 < μ1. The ray bends away from the normal. A convex boundary can behave as a diverging surface for this direction.
3
Cartesian Sign Convention
Always decide the direction of incident light first. Take the pole P as origin. Distances measured in the direction of incident light are positive; opposite distances are negative.
| Quantity | Positive | Negative |
|---|---|---|
| Object distance u | Object on outgoing side (virtual object) | Real object on incident side |
| Image distance v | Real image on outgoing side | Virtual image on incident side |
| Radius R | C on outgoing side of P | C on incident side of P |
| Height | Above principal axis | Below principal axis |
Rarer to denser: object in μ₁, image side μ₂
−μ1/u + μ2/v = (μ2 − μ1)/R
Denser to rarer: object in μ₂, image side μ₁
−μ2/u + μ1/v = (μ1 − μ2)/R
These are not different laws. They are the same general equation with the incident-medium index written first and the refracted-medium index written second. Use the first when light starts in μ1 and enters μ2; use the second when the direction is reversed.
4
Point-Object Angular Derivation Method
Every proof below starts from its own ray diagram. O and I are points on the principal axis. A is the point of incidence, and AC is the normal because C is the centre of the spherical surface.
α = angle made by incident ray OA with the principal axis
β = angle made by normal AC with the principal axis
γ = angle made by refracted ray AI with the principal axis
Paraxial approximationsin i ≈ i, sin r ≈ r, tan α ≈ α, tan β ≈ β, tan γ ≈ γ
Meaning of AP: AP is only the perpendicular offset of the incidence point A from the axis. It is not object height or image height. Since the same AP occurs in all three small-angle ratios, it cancels.
Fixed medium arrangement: μ1 is always drawn on the left and μ2 is always drawn on the right. Cases 3 and 5 reverse the direction of light; they do not interchange the media. The positive Cartesian direction is always the direction of incident light.
Read each picture independently. The relations among i, r, α, β and γ change when the image becomes virtual, the surface becomes concave, or the direction of refraction is reversed.
5
Five Separate Case-Wise Derivations
In all five diagrams μ1 remains on the left and μ2 remains on the right. Red solid lines are actual rays, red dotted lines are virtual extensions, and the green dashed line AC is the normal. Angle arcs are schematic and enlarged for clarity.
Case 1 · Convex Surface · Rarer → Denser · Real Image
μ2 > μ1; u < 0, v > 0 and R > 0.
Physical check: μ₂ > μ₁, so the refracted ray bends towards AC and r < i. The solid refracted ray actually reaches I in μ₂.
i = α + βr = β − γα ≈ AP/(−u), β ≈ AP/R, γ ≈ AP/v
Snell: μ1 sin i = μ2 sin r
Paraxial rays: μ1i = μ2r
μ1(α + β) = μ2(β − γ)
μ1α + μ2γ = (μ2 − μ1)β
μ1AP/(−u) + μ2AP/v = (μ2 − μ1)AP/R; cancel AP.
Case 1 result−μ1/u + μ2/v = (μ2 − μ1)/R
Case 2 · Convex Surface · Rarer → Denser · Virtual Image
The object is inside the first focal distance; u < 0, v < 0 and R > 0.
Physical check: μ₂ > μ₁, so r < i. The solid refracted ray does not pass through I; only its dotted backward extension does. I is on the object side and farther from P than the nearby object.
i = α + βr = β + γα ≈ AP/(−u), β ≈ AP/R, γ ≈ AP/(−v)
Snell: μ1 sin i = μ2 sin r
Paraxial rays: μ1i = μ2r
μ1(α + β) = μ2(β + γ)
μ1α − μ2γ = (μ2 − μ1)β
μ1AP/(−u) − μ2AP/(−v) = (μ2 − μ1)AP/R; cancel AP.
Case 2 result−μ1/u + μ2/v = (μ2 − μ1)/R
Case 3 · Convex Surface · Denser → Rarer · Real Image
μ₁ remains left and μ₂ remains right. O and C are in μ₂, and O is beyond the first focal distance for this direction. Light travels right-to-left: u < 0, v > 0 and R < 0.
Physical check: Light goes from μ₂ to μ₁, so it bends away from AC and r > i. The oblique refracted ray and the undeviated axial ray actually intersect at I in μ₁; therefore the image is real.
i = β − αr = β + γα ≈ AP/(−u), β ≈ AP/(−R), γ ≈ AP/v
Snell: μ2 sin i = μ1 sin r
Paraxial rays: μ2i = μ1r
μ2(β − α) = μ1(β + γ)
−μ2α − μ1γ = (μ1 − μ2)β
−μ2AP/(−u) − μ1AP/v = (μ1 − μ2)AP/(−R)
Cancel AP and multiply by −1. Since incident light travels right-to-left, u < 0, v > 0 and R < 0.
Case 3 result−μ2/u + μ1/v = (μ1 − μ2)/R
Case 4 · Concave Surface · Rarer → Denser
C is on the incident side, so R < 0. The diagram gives a virtual image with u < 0 and v < 0.
Physical check: Light enters the denser medium, so the refracted ray bends towards AC and r < i. Its dotted backward extension intersects the undeviated axial ray at I on the object side; therefore I is virtual.
i = β − αr = β − γα ≈ AP/(−u), β ≈ AP/(−R), γ ≈ AP/(−v)
Snell: μ1 sin i = μ2 sin r
Paraxial rays: μ1i = μ2r
μ1(β − α) = μ2(β − γ)
−μ1α + μ2γ = (μ2 − μ1)β
−μ1AP/(−u) + μ2AP/(−v) = (μ2 − μ1)AP/(−R)
Cancel AP: μ1/u − μ2/v = −(μ2 − μ1)/R. Multiplying by −1 gives the result.
Case 4 result−μ1/u + μ2/v = (μ2 − μ1)/R
Case 5 · Concave Surface · Denser → Rarer
The media stay fixed: μ₁ left, μ₂ right. O and virtual I are both in μ₂. Light travels right-to-left; u < 0, v < 0 and R > 0.
Physical check: Light leaves μ₂ for μ₁, so r > i. The actual ray travels into μ₁, but its dotted backward extension gives virtual I in μ₂ on the same side as O.
i = α + βr = β + γα ≈ AP/(−u), β ≈ AP/R, γ ≈ AP/(−v)
Snell: μ2 sin i = μ1 sin r
Paraxial rays: μ2i = μ1r
μ2(α + β) = μ1(β + γ)
μ2α − μ1γ = (μ1 − μ2)β
μ2AP/(−u) − μ1AP/(−v) = (μ1 − μ2)AP/R
Cancel AP. Here positive is right-to-left, so C is on the positive side while O and I are on the negative side.
Case 5 result−μ2/u + μ1/v = (μ1 − μ2)/R
6
Object at Infinity, Image at Infinity and Special Cases
Object at Infinity
Set u → −∞, so μ1/u → 0.
v = μ2R/(μ2 − μ1)
This v is the second focal length f2.
Image at Infinity
Set v → ∞, so μ2/v → 0.
u = −μ1R/(μ2 − μ1)
The magnitude gives the first focal length f1.
Plane surface
R → ∞ gives μ2/v = μ1/u.
Equal indices
μ1 = μ2 gives zero surface power and no bending.
Surface power
Φ = (μ2 − μ1)/R in m−1.
Focal relation
f2/f1 = −μ2/μ1 with signed focal lengths.
Lens-maker connection: Apply the spherical-surface formula successively at the two surfaces of a thin lens. With the thin-lens approximation, this leads to 1/f = (μ − 1)(1/R1 − 1/R2) for a lens in air.
7
Numerical Problems: Core Method
For every problem: sketch the direction of light, identify the incident and refracted media, assign signs, then substitute once.
1 · Given
Write μ1, μ2, u, R and required v.
2 · Signs
Use P as origin and the incident-light direction as positive.
3 · Formula
μ2/v − μ1/u = (μ2 − μ1)/R.
4 · Interpret
v > 0 means real; v < 0 means virtual.
8
Problems for Practice: Refraction at Spherical Surface
Open each solution to see the complete Given → Formula → Substitution → Calculation → Final Answer sequence.
1A spherical convex surface of radius of curvature 20 cm, made of glass (μ = 1.5) is placed in air. Find the position of the image formed, if a point object is placed at 30 cm in front of the convex surface on the principal axis. [CBSE SP 18]
View step-by-step solution
Givenμ1 = 1, μ2 = 1.5, u = −30 cm, R = +20 cm.
Formulaμ2/v − μ1/u = (μ2 − μ1)/R
Substitution1.5/v − 1/(−30) = (1.5 − 1)/20
Calculation1.5/v + 1/30 = 1/40 ⇒ 1.5/v = −1/120 ⇒ v = −180 cm.
Final AnswerThe negative sign places the image on the incident side.
Virtual image at 180 cm in air.
2A convex refracting surface of radius of curvature 20 cm separates two media of refractive indices 4/3 and 1.60. An object is placed in the first medium (μ = 4/3) at a distance of 200 cm from the refracting surface. Calculate the position of the image formed.
View step-by-step solution
Givenμ1 = 4/3, μ2 = 1.60, u = −200 cm, R = +20 cm.
Formulaμ2/v − μ1/u = (μ2 − μ1)/R
Substitution1.60/v − (4/3)/(−200) = (1.60 − 4/3)/20
Calculation1.60/v + 1/150 = 1/75 ⇒ 1.60/v = 1/150 ⇒ v = 240 cm.
Final Answerv is positive, so the rays meet in the second medium.
At 240 cm in denser medium.
3A concave spherical surface of refractive index 3/2 is immersed in water of refractive index 4/3. If a point object lies in water at a distance of 10 cm from the pole of the refracting surface, calculate the position of the image. Given radius of curvature = 18 cm.
View step-by-step solution
Givenμ1 = 4/3, μ2 = 3/2, u = −10 cm, R = −18 cm.
Formulaμ2/v − μ1/u = (μ2 − μ1)/R
Substitution(3/2)/v − (4/3)/(−10) = [(3/2) − (4/3)]/(−18)
Calculation1.5/v + 0.13333 = −0.009259 ⇒ 1.5/v = −0.142592 ⇒ v ≈ −10.52 cm.
Final AnswerThe negative sign means a virtual image in water.
Virtual image at 10.52 cm from the pole and inside the water.
4A mark placed on the surface of a glass sphere is viewed through glass from a position directly opposite. If the diameter of the sphere is 10 cm and refractive index of glass is 1.5, find the position of the image.
View step-by-step solution
GivenRadius = 5 cm; object mark is 10 cm from the viewing surface. For glass → air: μ1 = 1.5, μ2 = 1, u = −10 cm, R = −5 cm.
Formulaμ2/v − μ1/u = (μ2 − μ1)/R
Substitution1/v − 1.5/(−10) = (1 − 1.5)/(−5)
Calculation1/v + 0.15 = 0.10 ⇒ 1/v = −0.05 ⇒ v = −20 cm.
Final AnswerThe virtual image lies 20 cm from the viewing surface, towards the mark.
20 cm towards the mark from the surface opposite to the mark.
5A glass sphere of radius 15 cm has a small bubble 6 cm from its centre. The bubble is viewed along a diameter of the sphere from the side on which it lies. How far from the surface will it appear to be if refractive index of glass is 1.5?
View step-by-step solution
GivenActual depth from near surface = 15 − 6 = 9 cm. μ1 = 1.5, μ2 = 1, u = −9 cm, R = −15 cm.
Formulaμ2/v − μ1/u = (μ2 − μ1)/R
Substitution1/v − 1.5/(−9) = (1 − 1.5)/(−15)
Calculation1/v + 1/6 = 1/30 ⇒ 1/v = −2/15 ⇒ v = −7.5 cm.
Final AnswerThe virtual image is inside the sphere.
Virtual image is seen at 7.5 cm from the spherical surface.
6An object is placed 50 cm from the surface of a glass sphere of radius 10 cm along the diameter. Where will the final image be formed after refraction at both the surfaces? Refractive index of glass μ = 1.5.
View step-by-step solution
GivenSphere radius = 10 cm, diameter = 20 cm, μ = 1.5; first-surface u1 = −50 cm, R1 = +10 cm.
FormulaApply μ2/v − μ1/u = (μ2 − μ1)/R at each surface.
SubstitutionFirst: 1.5/v1 − 1/(−50) = 0.5/10 ⇒ v1 = 50 cm. This point is 30 cm beyond the second surface, so u2 = +30 cm and R2 = −10 cm.
CalculationSecond: 1/v2 − 1.5/30 = (1 − 1.5)/(−10) ⇒ 1/v2 = 0.10 ⇒ v2 = +10 cm from the second surface.
Final AnswerThe second surface is 10 cm to the right of the centre, so the image is 20 cm from the centre.
At 20 cm from the centre of the sphere.
7A spherical surface of radius 30 cm separates two transparent media A and B with refractive indices 1.33 and 1.48 respectively. Medium A is on the convex side of the surface. Where should a point object be placed in medium A so that the paraxial rays become parallel after refraction at the surface?
View step-by-step solution
Givenμ1 = 1.33, μ2 = 1.48, R = +30 cm and v = ∞.
Formula0 − μ1/u = (μ2 − μ1)/R
Substitution−1.33/u = (1.48 − 1.33)/30
Calculation−1.33/u = 0.15/30 = 0.005 ⇒ u = −266 cm.
Final AnswerThe object is on the incident side in medium A.
At 266 cm from the pole.
8A small air bubble is inside a glass sphere (μ = 1.5) of radius 10 cm. The bubble is 4.0 cm below the surface and is viewed normally from outside. Find the apparent depth of the bubble.
View step-by-step solution
GivenGlass → air: μ1 = 1.5, μ2 = 1, u = −4.0 cm, R = −10 cm.
Formulaμ2/v − μ1/u = (μ2 − μ1)/R
Substitution1/v − 1.5/(−4) = (1 − 1.5)/(−10)
Calculation1/v + 0.375 = 0.05 ⇒ 1/v = −0.325 ⇒ v ≈ −3.08 cm ≈ −3.0 cm.
Final AnswerThe negative sign denotes a virtual image inside the glass.
3 cm below the surface.
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PYQs and International Exam Practice
Short-answer practice
- Derive the spherical refraction formula using Snell’s law and the paraxial approximation.
- A convex glass-water boundary is reversed. Explain how its power and focal lengths change.
- Identify the signs of u, v and R for a virtual image produced by a concave boundary.
Multi-concept practice
- Two spherical refracting surfaces share an axis. Find the final image by using the first image as the second object.
- Find the surface radius required to convert a diverging pencil into a parallel beam.
- For a transparent sphere, determine the condition for the second surface to receive a virtual object.
Explanation and data response
- Explain why a bubble in a glass sphere appears closer to the observer.
- Use a scaled ray diagram to distinguish real rays from virtual extensions.
- Discuss the validity of the small-angle approximation and one source of uncertainty.
Reasoning-led questions
- Students vary object distance and graph μ2/v against μ1/u. Predict slope and intercept.
- Design a method to estimate the refractive index of a transparent sphere from apparent-depth data.
- Compare the image predicted by a plane-interface model with a spherical-interface model.
Answer hints for exam practice
Graph: Rearranging gives μ2/v = μ1/u + (μ2 − μ1)/R, so the graph has slope +1 and intercept equal to surface power.
Successive surfaces: Locate the first image in one coordinate system, then convert its position to u for the next pole. Never reuse the same origin.
Approximation: The theory is strongest for small aperture, rays close to the axis and angles sufficiently small that sin θ ≈ tan θ ≈ θ.
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Case Studies and Applications
1 · Apparent depth in spherical glass
Unlike a plane slab, curvature contributes optical power. Apparent depth depends on μ, object depth and R, not merely on the ratio of refractive indices.
2 · Bubble inside a glass sphere
The bubble is a real object for the glass-air boundary. Its virtual image generally appears closer to the nearest surface.
3 · Fish-eye / spherical water surface
A curved water-air boundary changes both apparent depth and angular field. The observed position differs from the flat-water estimate.
4 · Lens-maker formula connection
A lens is two spherical refractors in sequence. Combining their powers under the thin-lens approximation produces the lens-maker equation.
5 · Refraction through a transparent sphere
The image due to the first boundary becomes the object for the second. Its signed location must be measured from the second pole.
Experimental extension
Use a travelling microscope to measure actual and apparent positions, then fit the spherical-surface equation to estimate μ.
Σ
Formula Revision Sheet
General formulaμ2/v − μ1/u = (μ2 − μ1)/R
Surface powerΦ = (μ2 − μ1)/R
Second focal lengthf2 = μ2R/(μ2 − μ1)
First focal lengthf1 = −μ1R/(μ2 − μ1)
✓
Sign Table, Common Mistakes and Exam Tips
Quick sign table
- Real object: u < 0
- Real image: v > 0
- Virtual image: v < 0
- C right of P: R > 0
- C left of P: R < 0
Common mistakes
- Using the lens formula at one surface
- Forgetting to reverse media indices
- Giving R a sign from shape alone
- Measuring second-surface u from the first pole
- Calling every negative answer incorrect
Exam tips
- Draw a tiny axis sketch before calculation
- Write signs in the Given line
- Keep all lengths in one unit
- State real/virtual and the medium
- For spheres, solve one surface at a time
Quick revision: Incident index first, refracted index second; pole is the origin; direction of incident light is positive; C decides the sign of R; the sign of v decides whether the image is real or virtual.
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