Q.Light from a point source in air falls on a spherical glass surface (n=1.5 and radius of curvature =20 cm). The distance of the light source from the glass surface is 100 cm. At what position the image is formed?
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Refraction at a Spherical Surface
Imagine you're looking at a fish in a pond. The fish appears closer to the surface than it actually is. That's refraction — light bends when it moves from water to air. Now take that idea and replace the flat water surface with a curved one, like a glass lens or a drop of water. That's refraction at a spherical surface.
The core intuition
When light hits a flat surface (like a glass slab), it bends once and travels straight. But when the surface is curved — part of a sphere — the angle at which light hits changes depending on where on the surface it strikes. A ray hitting near the centre meets the surface almost head-on; a ray hitting near the edge meets it at a steep slant. This variation in incidence angle is what makes spherical surfaces focus or diverge light.
Think of a spherical surface as a tiny piece of a sphere. The centre of that sphere is called the centre of curvature (C). The distance from the surface to C is the radius of curvature (R). The line joining the centre of the surface (the pole, P) to C is the principal axis.
The precise geometry
We need to track what happens to a ray from an object point O on the principal axis. The ray travels in medium 1 (refractive index n1), hits the spherical surface at point A, and enters medium 2 (refractive index n2). The surface has radius R, with centre C.
The key is Snell's law at point A:
n1sini=n2sinr
But i and r are measured from the normal at A. For a spherical surface, the normal at any point is the line joining that point to C. So the normal is AC.
For small angles (paraxial rays — rays close to the axis), sinθ≈θ in radians. This approximation is the backbone of all standard lens and mirror formulas. It lets us replace Snell's law with:
n1i=n2r
Now look at the geometry. Let the object distance from the pole be u (negative by sign convention — object on left), and the image distance be v (positive if image is on the right, in medium 2). The angle the incident ray makes with the axis is α, the refracted ray makes β, and the normal makes θ with the axis.
From the triangles:
- In △OAC: i=α+θ
- In △AIC: r=θ−β (for a convex surface towards the object)
Substitute into Snell's law:
n1(α+θ)=n2(θ−β)
For small angles, α≈POAP≈−uh (since u is negative), β≈vh, and θ≈Rh.
Plugging these in:
n1(−uh+Rh)=n2(Rh−vh)
Cancel h (non-zero) and rearrange:
vn2−un1=Rn2−n1
This is the refraction at a spherical surface formula. It relates object distance u, image distance v, radii R, and the two refractive indices.
Sign convention (crucial for exams)
Use the Cartesian sign convention (the one used in NCERT and most Indian boards):
- Distances measured from the pole P along the principal axis.
- Positive in the direction of incident light (usually left to right).
- Negative opposite to incident light.
- R is positive if the centre of curvature C is on the right (convex surface towards object), negative if C is on the left (concave surface towards object).
The most common mistake is getting the sign of R wrong. Always check: is the centre of curvature on the same side as the incoming light or the opposite side? If opposite, R is positive.
What the formula tells you
- If n2>n1 (going from rarer to denser), the right side Rn2−n1 is positive for a convex surface. This means v is positive — the image forms on the other side (real image). …
Why this formula?
Great — let’s build the Refraction at a Spherical Surface formula from first principles. The goal is to understand why the relation
vn2−un1=Rn2−n1
holds, where:
- n1 = refractive index of the first medium (where the object lies)
- n2 = refractive index of the second medium (where the image lies)
- u = object distance from the pole (sign convention: negative for real object)
- v = image distance from the pole (sign convention: positive for real image on the opposite side)
- R = radius of curvature of the spherical surface (positive if centre of curvature is on the image side)
1. The core idea: Snell’s law at a curved interface
At any point on the spherical surface, the incident ray and refracted ray obey Snell’s law:
n1sini=n2sinr
For small angles (paraxial approximation — rays close to the principal axis), sinθ≈θ (in radians). So:
n1i=n2r
This linearisation is the key that lets us turn geometry into algebra.
2. Geometry of a single ray
Consider a point object O on the principal axis. A ray from O strikes the spherical surface at point P (height h above the axis). Let:
- C = centre of curvature of the spherical surface
- M = pole of the surface (vertex)
- I = image point formed after refraction
Draw the normal at P — it passes through C (since the surface is spherical). The angles:
- i = angle between incident ray OP and the normal PC
- r = angle between refracted ray PI and the normal PC
3. Relating angles to distances (paraxial approximation)
Because h is small compared to u, v, and R:
- Angle between OP and the axis: α≈uh (with sign)
- Angle between PC (normal) and the axis: θ≈Rh
- Angle between PI and the axis: β≈vh
Now, from the geometry of the triangle formed by the ray, the normal, and the axis:
- Incident angle i = angle between OP and the normal = θ−α (if θ>α)
- Refracted angle r = angle between PI and the normal = θ−β
Check the sign convention carefully — the exact relation depends on whether the ray bends toward or away from the normal. For a convex surface (centre on the image side), the standard result is:
i=α+θandr=θ−β
But the difference that matters is:
i−r=α+β
4. Applying Snell’s law
From n1i=n2r, we can write:
n1i=n2(i−(i−r))or directly:
n1i=n2r⟹n1i−n2r=0
But it’s more useful to express r in terms of i and the geometry:
r=i−(α+β)
Substitute into Snell’s law:
n1i=n2[i−(α+β)]
Simplify:
n1i=n2i−n2(α+β)
(n1−n2)i=−n2(α+β)
Now, i≈α+θ (from geometry). For small angles, α≈h/u, β≈h/v, θ≈h/R.
5. Substituting the small-angle approximations
Let’s do it step by step:
(n1−n2)(α+θ)=−n2(α+β)
Replace α, β, θ:
(n1−n2)(uh+Rh)=−n2(uh+vh)
Cancel h (non-zero):
(n1−n2)(u1+R1)=−n2(u1+v1)
6. Rearranging to the standard form
Expand the left side: …
Concept: Refraction at a single spherical surface.
Reasoning:
- Use the standard formula for refraction at a spherical surface:
vn2−un1=Rn2−n1
where $n_1 = 1$ (air), $n_2 = 1.5$ (glass), $u = -100\ \text{cm}$ (real object, negative sign convention), and $R = +20\ \text{cm}$ (convex surface towards the object).
2. Substitute the values:
v1.5−−1001=201.5−1
$$ \frac{1.5}{v} + \frac{1}{100} = \frac{0.5}{20} = \frac{1}{40} $$ …
Using the refraction formula for a single spherical surface, the image forms 100 cm behind the glass surface, on the opposite side from the object — a real image.
The problem is a classic application of refraction at a single spherical surface. Light travels from air (refractive index n1=1) into glass (n2=1.5). The surface is convex toward the incident light (the source is in air, the glass is on the other side). The radius of curvature R=+20 cm by the sign convention: the centre of curvature lies on the side where light goes (into the glass), so R is positive.
The key formula that governs this is the single spherical surface equation, which relates object distance u, image distance v, the two refractive indices, and the radius of curvature.
vn2−un1=Rn2−n1
Here, n1=1 (air), n2=1.5 (glass), R=+20 cm, and u=−100 cm (object is real, on the incident side, so u is negative by the Cartesian sign convention).
Let’s work through it step by step.
- Set up the equation with signs. The object is real and placed in front of the surface, so u=−100 cm. The surface is convex toward the object, so R=+20 cm. Substitute into the formula:
v1.5−−1001=201.5−1
- Simplify the right-hand side.
200.5=401
- Simplify the left-hand side. The term −−1001 becomes +1001:
v1.5+1001=401
- Isolate v1.5.
v1.5=401−1001
Find a common denominator (200):
401=2005,1001=2002
So:
v1.5=2005−2=2003
- Solve for v.
1.5×3200=v
v=3300=100 cm …
Method: Refraction Formula for a Single Spherical Surface
We use the Gaussian formula for refraction at a spherical surface:
vn2−un1=Rn2−n1
Step-by-step solution
Step 1: Identify the given values
- Refractive index of air, n1=1
- Refractive index of glass, n2=1.5
- Radius of curvature, R=+20 cm (convex surface toward the object in air)
- Object distance from surface, u=−100 cm (negative by sign convention — object is real and on the incident side)
Step 2: Apply the formula
v1.5−−1001=201.5−1
Step 3: Simplify
v1.5+1001=200.5
v1.5+0.01=0.025
Common Mistakes & How to Avoid Them: Refraction at Spherical Surface
Mistake 1: Wrong Sign Convention for R
The error: Students often take R=+20 cm without checking the surface orientation.
In this problem, light travels from air to glass through a convex surface (since the source is in air and the centre of curvature lies on the other side).
Correct approach:
- For a convex surface (centre of curvature on the transmitted side), R is positive.
- Here, R=+20 cm is correct — but only if you verify the direction.
How to avoid:
Always draw a ray diagram. Mark the centre of curvature C. If C lies on the side where light is going after refraction, R>0. If it lies on the incident side, R<0.
Mistake 2: Confusing n1 and n2 in the Formula
The error: Swapping refractive indices in the formula:
vn2−un1=Rn2−n1
Students sometimes write vn1−un2 or misplace n1 and n2.
Correct assignment:
- n1 = refractive index of the incident medium = 1 (air)
- n2 = refractive index of the transmitted medium = 1.5 (glass)
How to avoid:
Write the formula as:
vntransmitted−unincident=Rntransmitted−nincident
Always label n1 and n2 before plugging numbers.
Mistake 3: Forgetting the Sign of u
The error: Using u=+100 cm instead of u=−100 cm.
Correct approach:
By Cartesian sign convention, object distance u is negative when the object is on the incident side (real object).
So u=−100 cm.
How to avoid:
Remember: real object → u is negative. Only virtual objects (rare in basic problems) give positive u.
Mistake 4: Arithmetic Errors in Solving for v
The error: Rushing through the algebra and making sign mistakes.
Correct calculation:
v1.5−(−100)1=201.5−1
v1.5+1001=200.5=401
v1.5=401−1001=2005−2=2003
v=1.5×3200=3300=100 cm
How to avoid:
- Write each step clearly.
- Keep fractions — avoid decimals until the final step. …
- GUJCET 2026Set x1 markMCQQ.Two thin convex lenses of focal length f1 and f2 are placed in contact with each other. The equivalent power of the lens combination is ______. (A) f1×f2 (B) f1×f2f1+f2 (C) f1+f2 (D) f1+f2f1×f2
›Reveal solutionSolution
Two thin lenses in contact combine by adding powers: P=f1f2f1+f2.
For thin lenses in contact, the equivalent power is the sum of the individual powers:
P=P1+P2=f11+f21
Taking the common denominator: …
- GUJCET 2025Set 031 markMCQQ.What is the power of combination of convex lens and concave lens of equal focal length 25 cm? (A) Zero (B) 25D (C) Infinite (D) 8D
›Reveal solutionSolution
Equal-magnitude powers of opposite sign add to zero.
Concept — power of a lens combination. Power P=f1 (in metres), and powers of thin lenses in contact add: P=P1+P2.
Steps.
- Convex lens: f1=+25 cm=+0.25 m⇒P1=+4 D. …
- GUJCET 2024Set 131 markMCQQ.A rays coming from an object which is situated at ∞ distance in air and falls on a spherical glass surface (n=1.5). Then the distance of image will be ________. R is the radius of curvature of a spherical glass. (A) 1.5R (B) R (C) 3R (D) 2R
›Reveal solutionSolution
Using vn2−un1=Rn2−n1 with parallel incident rays gives v=3R.
Steps. For refraction at a spherical surface, vn2−un1=Rn2−n1. With n1=1, n2=1.5, u=∞: …
- GUJCET 2023Set 091 markMCQQ.A lens of power −4.0 Diopter. It means ______. (A) Concave lens of focal length −25.0 cm (B) Concave lens of focal length −0.25 cm (C) Convex lens of focal length +0.25 cm (D) Convex lens of focal length +25.0 cm
›Reveal solutionSolution
P=−4D ⇒f=1/P=−0.25 m =−25 cm, a concave lens.
Concept — power and focal length. Power (in dioptre) is the reciprocal of focal length in metres:
P=f(m)1⇒f=P1=−4.01=−0.25 m=−25 cm …
- GUJCET 2022Set 171 markMCQQ.Light from a point source in air falls on a spherical glass surface (n=1.5 and radius of curvature = 20 cm). The distance of the light source from the glass surface is 100 cm. Find the image distance. (A) −100 cm (B) −200 cm (C) 200 cm (D) 100 cm
›Reveal solutionSolution
Use vn2−un1=Rn2−n1.
Concept. Refraction at a single spherical surface (air → glass). Sign convention: u=−100 cm, R=+20 cm, n1=1, n2=1.5.
Steps.
- v1.5−−1001=201.5−1=0.025. …
- GUJCET 2022Set 171 markMCQQ.Double-convex lenses are to be manufactured from a glass of refractive index 1.55 with both faces of the same radius of curvature. What is the radius of curvature required if the focal length is to be 20 cm? (A) 44 cm (B) 2.2 cm (C) 22 cm (D) 4.4 cm
›Reveal solutionSolution
For a symmetric double-convex lens f1=(n−1)R2.
Concept. Lensmaker: f1=(n−1)(R11−R21). For a double-convex lens with equal radii, R1=+R, R2=−R, giving f1=(n−1)R2. …
- GUJCET 2022Set 171 markMCQQ.What is the focal length of a convex lens of focal length 30 cm in contact with a concave lens of focal length 10 cm? [Ignore thickness of lens] (A) −15 cm (B) −40 cm (C) −20 cm (D) −30 cm
›Reveal solutionSolution
Lenses in contact: f1=f11+f21.
Concept. Convex f1=+30 cm, concave f2=−10 cm.
Steps. …
- GUJCET 2021Set 151 markMCQQ.The radii of curvature of the faces of a double convex lens are 10 cm and 15 cm. Its focal length is 12 cm. What is the refractive index of material of lens? (A) 1.33 (B) 1.62 (C) 1.50 (D) 2.42
›Reveal solutionSolution
Apply the lensmaker's equation with R1=+10, R2=-15 for a double-convex lens.
Concept. f1=(μ−1)(R11−R21).
Solution. …
- GUJCET 2021Set 151 markMCQQ.Find equivalent focal length due to combination of two convex lens are in contact having a focal length both of them 30 cm. (A) 15 cm (B) 30 cm (C) 20 cm (D) 40 cm
›Reveal solutionSolution
Thin lenses in contact add reciprocals of focal lengths.
Concept. F1=f11+f21. …
- GUJCET 2020Set 071 markMCQQ.For glass lens f=+50 cm. Then power of lens is ______. (A) +2 D (B) -2 D (C) +0.02 D (D) -0.02 D
›Reveal solutionSolution
Power (in dioptres) is the reciprocal of focal length in metres.
Concept: P=f(m)1, with a converging (convex) lens having positive power. …
- GUJCET 2019Set 131 markMCQQ.The focal length of a thin lens made from the material of refractive index 1.5 is 15 cm. When it is placed in a liquid of refractive index 34, its focal length will be ...............cm. (A) 60 (B) 78.23 (C) 50 (D) 80.31
›Reveal solutionSolution
The lens power scales with (nrel−1); moving from air to the denser liquid cuts the relative index term to one-fourth, so f grows 4x to 60 cm.
Concept: The lensmaker's equation gives f1=(nrel−1)(R11−R21), where nrel is the lens index relative to the surrounding medium. The bracket (geometry) is fixed; only nrel changes.
Steps: …
- GUJCET 2014Set A1 markMCQQ.A plano convex lens fits exactly into plano concave lens as shown in figure. Their plane surfaces are parallel to each other. If the lens are made of different materials of refractive indices 1.6 & 1.5 respectively. If R is the radius of curvature of curved surfaces of lenses. Then the focal length of the combination. [FIGURE] (A) 6.2R (B) 0.2R (C) 3.1R (D) 0.1R
›Reveal solutionSolution
[!TLDR]
Adding the two lens powers gives f1=R0.6−R0.5=R0.1, so f=0.1R; option (D).
Concept
For thin lenses in contact the net power is the sum of individual powers, f1=f11+f21. For a plano-curved lens only the single curved surface (radius R) contributes, so f1=R(n−1) in magnitude, positive for the convex piece and negative for the concave piece.
Solution
- Plano-convex, n=1.6: f11=R1.6−1=R0.6 (converging). …
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