A thin lens is placed on the axis between a point source S and an observer O. The source S lies on the axis a

NCERT Exemplar · Q32

u

(i) Let the thickness vary as

w(b)=w_0-\dfrac{b^2}{\alpha}

, where

w_0

and

\alpha

are constants. Using Fermat's principle (that the transit time of a ray from S to O is an extremum), find the condition under which all paraxial rays leaving S converge to the single point O on the axis, and hence find the focal length of the lens.

(ii) A gravitational lens may be modelled with a thickness profile

w(b)=k_1\ln\!\big(k_2/b\big)

for

b_{\min}<b<b_{\max}

(and constant, equal to

k_1\ln(k_2/b_{\min})

, for

b<b_{\min}

). Show that an observer sees the image of a point object as a ring about the centre of the lens, of angular radius

\beta=\sqrt{\dfrac{(n-1)\,k_1\,(u/v)}{u+v}}

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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 $\frac{n-1}{R}$ . 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.

Note

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:

\frac{1}{f} = (n-1)\left(\frac{1}{R_1} - \frac{1}{R_2}\right)

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)
$R_1$ is the radius of curvature of the first surface (the one light reaches first)
$R_2$ is the radius of curvature of the second surface

\frac{1}{f} = (n-1)\left(\frac{1}{R_1} - \frac{1}{R_2}\right)

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, $R_1 = +20$ 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 $R_2 = -20$ cm.

Plug in:

\frac{1}{f} = (1.5 - 1)\left(\frac{1}{20} - \frac{1}{-20}\right) = 0.5 \times \left(\frac{1}{20} + \frac{1}{20}\right) = 0.5 \times \frac{2}{20} = \frac{1}{20}

So $f = +20$ cm. Positive means converging — correct for a biconvex lens.

Watch out

The most common mistake is getting the sign of $R_2$ wrong. For a biconvex lens, $R_1$ is positive and $R_2$ is negative. For a biconcave lens, it's the reverse: $R_1$ negative, $R_2$ positive. Always sketch the lens and mark where each surface's centre of curvature actually sits.

Why the Formula Works (Brief Derivation) …

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Fermat's principle says a physically realised ray makes the optical path length $L(b)$ stationary. If $L(b)$ can be made independent of $b$ , every paraxial ray takes the same time and they all meet at O. Imposing this on $w(b)=w_0-b^2/\alpha$ fixes the lens equation and the focal length. For the logarithmic (gravitational-lens) width, $L(b)$ is stationary at a single radius $b$ , and by circular symmetry the image is a ring; its angular radius reduces to the quoted $\beta$ .

Optical path length

A ray from S crossing the lens at height $b$ travels $\sqrt{u^2+b^2}$ to the lens, then $\sqrt{v^2+b^2}$ to O, and in passing through the glass of thickness $w(b)$ it acquires an extra optical path $(n-1)w(b)$ relative to air. In the paraxial limit $b\ll u,v$ ,

L(b)\approx u+v+\frac{b^2}{2u}+\frac{b^2}{2v}+(n-1)\,w(b).

Part (i): $w(b)=w_0-\dfrac{b^2}{\alpha}$

Substituting,

L(b)=\big(u+v+(n-1)w_0\big)+\frac{b^2}{2}\!\left(\frac1u+\frac1v\right)-(n-1)\frac{b^2}{\alpha}.

For all paraxial rays to converge at O, $L(b)$ must not depend on $b$ , so the coefficient of $b^2$ must vanish:

\frac12\!\left(\frac1u+\frac1v\right)-\frac{(n-1)}{\alpha}=0\;\;\Longrightarrow\;\;\boxed{\frac1u+\frac1v=\frac{2(n-1)}{\alpha}}.

Comparing with the thin-lens relation $\dfrac1u+\dfrac1v=\dfrac1f$ ,

\boxed{\,f=\frac{\alpha}{2(n-1)}\,}.

Part (ii): $w(b)=k_1\ln\!\big(k_2/b\big)$

Now

L(b)=u+v+\frac{b^2}{2}\!\left(\frac1u+\frac1v\right)+(n-1)k_1\ln\!\frac{k_2}{b}.

Apply Fermat's principle, $\dfrac{\mathrm dL}{\mathrm db}=0$ :

b\!\left(\frac1u+\frac1v\right)+(n-1)k_1\!\left(-\frac1b\right)=0\;\;\Longrightarrow\;\;b^2\!\left(\frac1u+\frac1v\right)=(n-1)k_1.

Since $\dfrac1u+\dfrac1v=\dfrac{u+v}{uv}$ ,

$b^2=(n-1)k_1\,\frac{uv}{u+v}.$ …

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Method: Using Fermat's principle to derive a lens condition from optical path length

This method applies to any problem asking you to derive a focusing condition (a lens equation, a focal length, an image location or shape) from a stated thickness/shape profile of a refracting element, using Fermat's principle rather than ray-by-ray Snell's-law tracing.

Steps

Step 1: Write the optical path length $L$ as a function of the transverse ray coordinate.

For a ray leaving a source, crossing the element at transverse height $b$ , and arriving at an image/observer point, the optical path length is the sum of the geometric path lengths in air plus the extra optical path picked up by travelling through the extra thickness of the (higher-index) medium:

L(b) = (\text{geometric path length through air}) + (n-1)\,w(b),

where $w(b)$ is the thickness of the medium at height $b$ and $(n-1)w(b)$ is the standard "extra optical path due to slowing down in glass" term. In the paraxial approximation, expand the geometric distances (each of the form $\sqrt{d^2+b^2}$ ) to leading order in $b$ :

\sqrt{d^2+b^2}\approx d+\frac{b^2}{2d}.

Step 2: Decide what Fermat's principle demands for this physical situation.

Fermat's principle always says the physically realised path makes $L(b)$ stationary, $\dfrac{dL}{db}=0$ . There are two distinct ways this gets used, and you must identify which one the question needs:

If every paraxial ray (every $b$ ) must reach the same image point — a true focus — then $L(b)$ must be independent of $b$ entirely, not merely stationary at one value. This means every coefficient of $b^2, b^4,\dots$ in the paraxial expansion must vanish.
If only a single ray (one particular $b$ ) is picked out — e.g. because the profile isn't the special "perfect lens" shape — then you literally solve $\dfrac{dL}{db}=0$ for the one value of $b$ that satisfies it, and that ray is the one that reaches the target.

Step 3: Solve the resulting algebraic equation for the quantity asked for. …

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Optical path length

Part (i): w(b)=w0−b2αw(b)=w_0-\dfrac{b^2}{\alpha}w(b)=w0​−αb2​

Part (ii): w(b)=k1ln⁡ ⁣(k2/b)w(b)=k_1\ln\!\big(k_2/b\big)w(b)=k1​ln(k2​/b)

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Part (i): $w(b)=w_0-\dfrac{b^2}{\alpha}$

Part (ii): $w(b)=k_1\ln\!\big(k_2/b\big)$

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