A thin convex lens of focal length 25 cm is cut into two pieces 0.5 cm above the principal axis. The top part is placed at (0,0) and an object placed at ($-50$ cm, 0). Find the coordinates of the image.

A thin convex lens of focal length 25 cm is cut into two pieces 0.5 cm above the principal axis. The top part

NCERT Exemplar · Q24

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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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The top half-lens still has $f = 25\ \text{cm}$ , but its optical axis is displaced $0.5\ \text{cm}$ below the origin. The lens formula gives $v = +50\ \text{cm}$ with $m = -1$ ; combining the $0.5\ \text{cm}$ axis shift with the inversion places the image at $(50\ \text{cm},\ -1\ \text{cm})$ .

Effect of the cut

Cutting a thin lens along a plane parallel to the principal axis does not change its focal length — each piece is still bounded by the same curved surfaces. What changes is geometry: the principal axis of the full lens passed through its centre, i.e. $0.5\ \text{cm}$ below the cut edge. When the top piece is placed with that cut edge at the origin, its optical axis lies at $y = -0.5\ \text{cm}$ .

Image distance

Measuring along the axis, $u = -50\ \text{cm}$ and $f = +25\ \text{cm}$ :

\frac{1}{v} - \frac{1}{u} = \frac{1}{f} \Rightarrow \frac{1}{v} = \frac{1}{25} + \frac{1}{-50} = \frac{2-1}{50} = \frac{1}{50},

so $v = +50\ \text{cm}$ — a real image $50\ \text{cm}$ to the right. The magnification is

$m = \frac{v}{u} = \frac{50}{-50} = -1.$ …

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Same / Similar Concept — real previous-year questions on the same or a closely similar concept, not this exact question.

Higher Secondary (+2 Stage) Examination 2026Set ANNUAL1 markMCQ
$Q.The lens shown in the figure is made of two different materials having refractive indices n1 and n2. If a point-sized object is placed on the axis of the lens, how many images will be formed?$
(a) 1
(b) 2
(c) 3
(d) 5
›Reveal solutionSolution

A lens split into two halves of different refractive index behaves like two lenses of different focal length sharing the same curvature — so a single point object produces two distinct images, one from each half.

When a lens is made of two materials of refractive indices n1 and n2 (split by a plane containing the principal axis), each half has the same radius of curvature but a different refractive index, so by the lens-maker's formula

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

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Higher Secondary (+2 Stage) Examination 2024Set ANNUAL1 mark
$Q. If red light is used instead of blue light, what change will occur in the focal length of a lens?$
›Reveal solutionSolution

Because glass bends red light less than blue light (lower refractive index for longer wavelengths), a lens has a slightly longer focal length for red light than for blue.

By the lens maker's formula:

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

Due to dispersion, the refractive index of glass is slightly higher for blue light (shorter wavelength) than for red light (longer wavelength), i.e. $n_{blue} > n_{red}$ . Since $\dfrac{1}{f} \propto (n-1)$ , a smaller $n$ (red light) gives a smaller $\dfrac1f$ , i.e. a LARGER focal length.

…

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Q.A thin convex lens of focal length 25 cm is cut into two pieces 0.5 cm above the principal axis. The top part is placed at (0,0) and an object placed at ( $-50$ cm, 0). Find the coordinates of the image.

Concept understanding — Lens Maker's Formula

The Intuition: Why a Lens Bends Light

The Precise Statement

How to Apply It: A Worked Example

Why the Formula Works (Brief Derivation) …

Effect of the cut

Image distance

Method: Handling a Decentred Optical Element (a Cut or Off-Axis Lens)

Steps

Q.A thin convex lens of focal length 25 cm is cut into two pieces 0.5 cm above the principal axis. The top part is placed at (0,0) and an object placed at (−50-50−50 cm, 0). Find the coordinates of the image.

Effect of the cut

Image distance

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Q.A thin convex lens of focal length 25 cm is cut into two pieces 0.5 cm above the principal axis. The top part is placed at (0,0) and an object placed at ( $-50$ cm, 0). Find the coordinates of the image.

Unlock everything free for 14 days