Q.The image of a small electric bulb fixed on the wall of a room is to be obtained on the opposite wall 3 m away by means of a large convex lens. What is the maximum possible focal length of the lens required for the purpose?
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🔒 Start your 14-day free trial to unlock the full solution →Concept understanding — Lens Maker's Formula
The Intuition: Why a Lens Bends Light
A lens works because light slows down when it enters glass. When a wavefront hits a curved surface at an angle, different parts of it slow down at different moments, and the wavefront bends. The stronger the curvature, the more it bends.
A lens has two surfaces. Each surface bends light by an amount that depends on its radius of curvature R and the refractive index n of the glass. The net bending — the focal length f — is the combined effect of both surfaces.
If you had a single spherical surface separating air from glass, its contribution to bending power is Rn−1. A lens has two such surfaces: light goes from air into glass at the first surface, then from glass back into air at the second. Because the two surfaces face opposite directions relative to the travelling light, their radii typically carry opposite signs.
This uses the New Cartesian Sign Convention (the one used in NCERT and CBSE): all distances are measured from the optical centre, and the direction the incident light travels in is taken as positive. So R is positive if the centre of curvature lies on the side the light is travelling towards (the outgoing side), and negative if it lies on the side the light is travelling from (the incident side).
The Precise Statement
For a thin lens (thickness negligible compared to the radii), the Lens Maker's Formula is:
f1=(n−1)(R11−R21)
where:
- f is the focal length of the lens (positive for converging, negative for diverging)
- n is the refractive index of the lens material relative to the surrounding medium (usually air)
- R1 is the radius of curvature of the first surface (the one light reaches first)
- R2 is the radius of curvature of the second surface
f1=(n−1)(R11−R21)
How to Apply It: A Worked Example
Take a biconvex lens made of glass (n=1.5) with both surfaces having the same radius of curvature magnitude, 20 cm.
Light travels left to right. The first surface bulges toward the incoming light, so its centre of curvature lies to the right of the surface — on the side the light is travelling towards. By the rule above, R1=+20 cm.
The second surface also bulges outward (away from the lens), so its centre of curvature lies to the left of that surface — on the side the light is travelling from. So R2=−20 cm.
Plug in:
f1=(1.5−1)(201−−201)=0.5×(201+201)=0.5×202=201
So f=+20 cm. Positive means converging — correct for a biconvex lens.
The most common mistake is getting the sign of R2 wrong. For a biconvex lens, R1 is positive and R2 is negative. For a biconcave lens, it's the reverse: R1 negative, R2 positive. Always sketch the lens and mark where each surface's centre of curvature actually sits.
Why the Formula Works (Brief Derivation) …
The key idea is that for a real image to be formed on a screen, the object and image distances must be positive. The lens formula gives a constraint on the focal length when the total distance between object and image is fixed.
Reasoning:
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Let the distance between the object (bulb) and the image (opposite wall) be D=3 m. If the lens is placed at a distance u from the object, then the image distance is v=D−u.
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Using the lens formula f1=u1+v1, we get:
f1=u1+D−u1=u(D−u)D. …
For a real image of an object on a wall to be formed on the opposite wall 3 m away using a single convex lens, the lens must be placed between them. The maximum focal length occurs when the object and image distances are equal, giving fmax=0.75 m.
The problem is about forming a real image of a real object using a convex lens. The object (the bulb) is fixed on one wall, and the image must be formed on the opposite wall, which is 3 m away. So the total distance between object and image is fixed at 3 m. The lens can be placed anywhere between them.
The key idea: For a convex lens, a real image is formed on the opposite side of the lens from the object. The lens formula is
f1=v1−u1
with the sign convention where u is negative (object distance measured from lens, opposite to incident light direction) and v is positive (real image on the other side). But it's simpler here to use magnitudes: let u be the distance from lens to object (positive), and v be the distance from lens to image (positive). Then the lens formula becomes
f1=u1+v1
and the total distance u+v=3 m.
We want the maximum possible focal length f for which a real image can be formed. Let's work through it.
- Set up the relation. Given u+v=3, we have v=3−u. Substitute into the lens formula:
f1=u1+3−u1
This is valid only when u>0 and v>0, i.e., 0<u<3.
- Express f as a function of u. Combine the fractions:
f1=u(3−u)3−u+u=u(3−u)3
So
f=3u(3−u)
This is a quadratic in u: f=31(3u−u2).
- Find the maximum of f. The expression u(3−u) is a downward-opening parabola in u, with maximum at the vertex. For a quadratic au2+bu+c, the vertex is at u=−2ab. Here u(3−u)=−u2+3u, so a=−1, b=3, giving …
Method: Lens Formula with Real Object and Real Image Condition
This problem uses the Lens Formula approach, combined with the constraint that both object and image are real (on opposite walls).
Step-by-step solution
Step 1: Identify the given data
- Distance between object and image: D=3 m
- Both object and image are real (on opposite walls)
- Lens is convex (converging)
Step 2: Apply the lens formula
The lens formula is:
f1=v1−u1
Where:
- u = object distance (negative by sign convention for real object)
- v = image distance (positive for real image)
Step 3: Express the distance constraint
If the lens is placed between the walls, and total distance is D:
v−u=D(since u is negative, -u is positive)
Let u=−x (where x>0 is the distance from lens to object wall). Then:
v=D−x
Step 4: Substitute into lens formula
f1=D−x1−−x1=D−x1+x1
Step 5: Find condition for maximum focal length
For a real image to form, we need f to be real and positive. The focal length is maximum when the denominator is minimum.
Rewrite:
f1=x(D−x)x+(D−x)=x(D−x)D
So:
f=Dx(D−x) …
Here are the common mistakes students make on this classic refraction/lens problem, and how to avoid each.
Mistake 1: Forgetting the Lens Formula Sign Convention
The error:
Students plug u=−3 m and v=+3 m directly into the lens formula without realising that the object and image are on opposite sides of the lens. They often treat both distances as positive, leading to a wrong focal length.
Why it happens:
They confuse the real object / real image situation with a simple "distance between object and image" approach.
How to avoid:
Always use the Cartesian sign convention:
- Object distance u is negative (real object, left of lens).
- Image distance v is positive (real image, right of lens).
So here:
- u=−x (distance from lens to object)
- v=+(3−x) (distance from lens to image, since total wall-to-wall distance is 3 m)
Mistake 2: Assuming Object and Image are at Equal Distances
The error:
Many students assume ∣u∣=∣v∣=1.5 m because the total distance is 3 m. They then compute f=0.75 m and stop.
Why it happens:
They think "symmetry" always gives the maximum focal length — but this is only true if the lens can be placed anywhere.
How to avoid:
Realise that the maximum focal length occurs when the lens is placed exactly midway only if you want a specific magnification. For maximum f, you must vary the lens position and find the condition where f is largest.
Mistake 3: Not Using the Condition for Real Image Formation
The error:
Students forget that for a real image, the lens formula must yield a positive f, and the quadratic in x must have real roots.
Why it happens:
They treat the problem as a simple substitution rather than an optimisation.
How to avoid:
Write the lens formula:
f1=v1−u1
Substitute u=−x, v=3−x:
f1=3−x1+x1
Then find the minimum of f1 (which gives maximum f) by differentiating or using AM–GM inequality.
Mistake 4: Algebraic Errors in the Quadratic
The error:
When simplifying f1=x(3−x)3, students often write f=3x(3−x) but then forget that f is maximum when the denominator x(3−x) is maximum. …
- JKBOSE Class 12 Annual Regular Examination 2026Set SZ5 marksQ.What is Lens Maker's formula? Derive an expression for Lens Maker's formula for a convex lens. OR State Huygen's Principle. Derive laws of reflection from Huygen's Principle.
›Reveal solutionSolution
The Lens Maker's formula, f1=(μ−1)(R11−R21), is derived by applying single-surface refraction twice, once at each face of the lens. (OR alternative: Huygens' Principle constructs wavefronts from secondary wavelets, and can be used to derive the laws of reflection geometrically.)
Part 1: Lens Maker's Formula
What it is. The Lens Maker's formula relates a lens's focal length f to the refractive index μ of its material (relative to the surrounding medium) and the radii of curvature R1, R2 of its two spherical surfaces — it tells a lens manufacturer what curvatures are needed to grind a lens of a desired focal length.
Derivation for a thin convex lens. Consider a thin lens with two refracting surfaces of radii R1 (first surface, light hits this first) and R2 (second surface), made of material of refractive index μ, surrounded by air (index 1). Let a point object O on the principal axis form an image after refraction at each surface in turn.
Step 1 — Refraction at the first surface (radius R1), treating it alone (ignoring the second surface for now), forming a virtual intermediate image I1 at distance v1:
v1μ−u1=R1μ−1
Step 2 — Refraction at the second surface (radius R2): the image I1 from step 1 now acts as the object for this second refraction (light going from the denser lens medium μ back into air, index 1), forming the final image I at distance v:
v1−v1μ=R21−μ
Step 3 — Add the two equations (the μ/v1 terms cancel):
v1−u1=(μ−1)(R11−R21)
Step 4 — Apply the lens definition. When the object is at infinity (u→∞), the image forms at the focus (v=f), so v1−u1→f1. Substituting:
f1=(μ−1)(R11−R21)
This is the Lens Maker's formula, and it also leads to the general thin lens formula v1−u1=f1.
OR: Huygens' Principle and the Laws of Reflection
Huygens' Principle. Every point on a given wavefront (a surface of constant phase) acts as a source of new secondary wavelets, which spread out in all directions with the speed of the wave in that medium. The new wavefront at any later time is given by the forward "envelope" (common tangent surface) of all these secondary wavelets.
Deriving the laws of reflection. Consider a plane wavefront AB incident on a reflecting surface MN at angle of incidence i, with A striking the surface first while B is still travelling. …
- JKBOSE Class 12 Annual Regular Examination 2023Set ANNUAL5 marksQ.Stating the assumptions made and convention of signs used, derive the lens maker's formula in case of a double convex lens. OR Define fringe width. Derive an expression for fringe width in Young's double slit experiment of interference of light.
›Reveal solutionSolution
The lens maker's formula, 1/f = (n21-1)(1/R1 - 1/R2), is derived by applying refraction at each of the two spherical surfaces of a thin lens in turn. (OR: Fringe width beta = lambdaD/d is derived from the path-difference condition for constructive/destructive interference in Young's double-slit experiment.)
Derivation of Lens Maker's Formula (for a thin double convex lens):
Assumptions: (i) The lens is thin, so the two refracting surfaces are close enough that the lateral displacement of the ray inside the lens can be neglected. (ii) Only paraxial rays (making small angles with the principal axis) are considered. (iii) The medium on both sides of the lens is the same (say, refractive index n1), and the lens material has refractive index n2.
Sign convention: All distances are measured from the pole/optical centre of the surface; distances measured in the direction of the incident light are taken positive, against it negative. For a double convex lens, the first surface (facing the incident light) is convex towards the object, so its radius R1 is positive; the second surface curves the other way, so its radius R2 is negative.
Consider a point object O on the principal axis. Let the first surface (radius R1) refract the light from the object; treating this surface alone, the image I1 formed (a virtual, intermediate image) obeys the single-surface refraction formula:
n2/v1 - n1/u = (n2 - n1)/R1 ... (i)
where u is the object distance and v1 is the image distance for the first surface alone.
This intermediate image I1 now acts as a virtual object for the second surface (radius R2), which refracts the ray back into the surrounding medium n1, forming the final real image I at distance v:
n1/v - n2/v1 = (n1 - n2)/R2 ... (ii)
Adding equations (i) and (ii), the n2/v1 and -n2/v1 terms cancel:
n1/v - n1/u = (n2-n1)/R1 + (n1-n2)/R2 = (n2-n1)*(1/R1 - 1/R2).
Dividing throughout by n1:
1/v - 1/u = (n2/n1 - 1)(1/R1 - 1/R2) = (n21 - 1)(1/R1 - 1/R2), where n21 = n2/n1 is the refractive index of the lens material relative to the surrounding medium.
Now, if the object is placed at infinity (u tends to infinity), the rays refracted by the lens converge (for a convex lens) to the principal focus, so v = f (the focal length). Putting u = infinity, 1/u = 0:
1/f = (n21 - 1)*(1/R1 - 1/R2).
This is the Lens Maker's Formula. It relates the focal length f of a thin lens to the refractive index of its material (relative to the surrounding medium) and the radii of curvature of its two surfaces, and is used by lens manufacturers to design a lens of a required focal length. For a double convex lens (R1 positive, R2 negative), (1/R1 - 1/R2) is positive, so f comes out positive - confirming it is a converging lens.
…
- JKBOSE Class 12 Annual Regular Examination 2021Set SZ5 marksQ.Derive Lens-Maker's formula for convex lens. Write the necessary sign convention used. OR State Huygen's wave principles. Use them to prove laws of refraction of light.
›Reveal solutionSolution
The lens-maker's formula relates a thin lens's focal length to its refractive index and the radii of curvature of its two surfaces, derived by applying single-surface refraction twice.
Sign convention (Cartesian, as used in NCERT): All distances are measured from the optical centre of the lens. Distances measured in the direction of the incident light are taken as positive; distances measured against the direction of incident light are taken as negative. Heights measured upward from the principal axis are positive, downward are negative. For a convex lens, if the centre of curvature of a surface lies on the outgoing-light side, its radius R is positive; if on the incoming-light side, R is negative.
Derivation of lens-maker's formula:
Consider a thin convex lens of refractive index n2 placed in a medium of refractive index n1, with surfaces of radii R1 and R2. Let an object be at O on the principal axis.
Refraction at the first surface (radius R1) forms an image at I1 (treating the second surface as absent), using the single spherical refracting surface formula:
v1n2−un1=R1n2−n1
Refraction at the second surface (radius R2): the image I1 from the first surface now acts as a virtual object for the second surface, forming the final image at I (at distance v):
vn1−v1n2=R2n1−n2
Adding these two equations (the n2/v1 terms cancel):
vn1−un1=(n2−n1)(R11−R21)
Dividing throughout by n1:
v1−u1=(n1n2−1)(R11−R21)=(n21−1)(R11−R21)
When the object is at infinity (u→∞), the image forms at the focus, v=f, giving the general lens formula v1−u1=f1, so:
f1=(n21−1)(R11−R21)
where n21=n2/n1 is the refractive index of the lens material relative to the surrounding medium.
OR — Huygens' principle and laws of refraction:
Huygens' wave principle: Every point on a given wavefront (locus of points vibrating in phase) acts as a source of new secondary wavelets, which spread out in all directions with the speed of the wave in that medium. The new (secondary) wavefront at any later instant is the surface tangent (envelope) to all these secondary wavelets.
Derivation of Snell's law using Huygens' construction:
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