JEE Mains Physics · Ray Optics

Thin Lens Formula and Lens Combinations

A thin lens images by 1/v − 1/u = 1/f with m = v/u; lenses in contact add their powers, and lenses apart are solved one after another, each image becoming the next lens's object.

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Why this matters

Twenty-six PYQs, fourteen of them multiple choice and twelve asking for a number, and five from 2026. Twelve use one lens: the distance between object and image with the magnification given, equal image sizes at two positions, a change in power, and readings or graphs from a focal-length experiment. Fourteen combine lenses: in contact, with a gap between them, as a pair that turns a parallel beam into a parallel beam, or with a mirror behind the lens.

Concept 1 of 2: Thin lens formula and magnification

The lens formula has a minus sign, 1/v − 1/u = 1/f, and the magnification has none, m = v/u. A convex lens has f > 0. With a real object it gives a real, inverted image when the object is beyond F, and a virtual, erect, magnified image when the object is inside F. A concave lens (f < 0) always gives a virtual, erect, diminished image.

Definition

$\dfrac{1}{v} - \dfrac{1}{u} = \dfrac{1}{f}$ , $m = \dfrac{v}{u} = \dfrac{f}{f + u}$ , $P = \dfrac{1}{f}$ (f in metres, P in dioptres).
Convex lens: object beyond 2F gives a real, inverted, diminished image between F and 2F; object between F and 2F gives a real, inverted, magnified image beyond 2F; object inside F gives a virtual, erect, magnified image on the object's side.
Concave lens: the image of a real object is always virtual, erect and diminished, between the lens and F.
Given m and the object-image distance: write $v = mu$ . A real image is on the far side, so the two distances add; a virtual image is on the object's side, so they subtract.
Equal image sizes at two object positions: one image is real, one virtual. Solve $\dfrac{f}{f + u_1} = -\dfrac{f}{f + u_2}$ .
Distances $x_1$ , $x_2$ of object and image from the two foci: $x_1 x_2 = f^{2}$ (Newton's form).
A plot of $1/|v|$ against $1/|u|$ for real images is a straight line that cuts both axes at $1/f$ .
A small change in power: $\dfrac{\Delta f}{f} = -\dfrac{\Delta P}{P}$ .

Thin lens formula

\frac{1}{v} - \frac{1}{u} = \frac{1}{f}, \qquad m = \frac{v}{u} = \frac{f}{f + u}, \qquad P = \frac{1}{f}

Worked example

A convex lens throws a real image, four times the size of the object, on a screen 75 cm from the object. (a) Find the focal length. (b) Where must the object be placed for an erect image four times as tall?

Practice this conceptself-check · 4 quick reps

The same idea in a real exam question:

JEE Mains · 2024 · 1 February 2024 · Q23Moderate

Example 1 · Ray Optics · Thin Lens Formula and Lens Combinations

The distance between object and its 3 times magnified virtual image as produced by a convex lens is

20\text{ }cm

. The focal length of the lens used is ____

cm

The lens formula has a minus sign

A lens uses 1/v − 1/u = 1/f and a mirror 1/v + 1/u = 1/f. Swapping them gives the wrong sign, or a wrong size altogether.

A concave lens never forms a real image of a real object

With f < 0 and u < 0, 1/v = 1/f + 1/u is negative: the image is always virtual, erect and inside F. A statement placing it at a real point on the far side is false.

A long or slanted object needs two magnifications

Heights across the axis scale by m; lengths along it scale by about m² only when the object is short. For anything longer, find the image of each end.

Drill 11 more on thin lens formula and magnification

Concept 2 of 2: Combinations of lenses

Lenses in contact act as one lens: their powers add. Lenses apart are solved one at a time. The image formed by the first lens is the object for the second, measured from the second lens. If that image lies beyond the second lens, the light is still converging when it arrives: it is a virtual object, with u > 0.

Definition

In contact: $P = P_1 + P_2 + \dots$ , that is $\dfrac{1}{F} = \dfrac{1}{f_1} + \dfrac{1}{f_2}$ . A convex and a concave lens in contact converge only if the convex one is the more powerful (shorter focal length).
Separated by d, as one equivalent lens: $P = P_1 + P_2 - dP_1P_2$ .
Step by step: image through lens 1, subtract the separation to get u for lens 2, image again. Total magnification $m = m_1 m_2$ .
Parallel beam in, parallel beam out (a telescope or a beam expander): two converging lenses are $f_1 + f_2$ apart, and the beam's width is multiplied by $f_2/f_1$ .
A lens followed by a curved mirror: the final image falls back on the object when the lens's image sits at the mirror's centre of curvature. Rays aimed at the centre meet the mirror normally and retrace their path.
A plane mirror behind a lens forms an image of the lens's image, as far behind the mirror as that image is in front of it.

Lenses in combination

P = P_1 + P_2 - dP_1P_2, \qquad m = m_1 m_2

Worked example

Two convex lenses of focal lengths 10 cm and 15 cm are 20 cm apart. An object is 15 cm in front of the first lens. (a) Find the final image and the total magnification. (b) What would the focal length be if the lenses were in contact?

Practice this conceptself-check · 4 quick reps

The same idea in a real exam question:

JEE Mains · 2026 · 22 Jan 2026 Shift 1 · Q14Moderate

Example 2 · Ray Optics · Thin Lens Formula and Lens Combinations

A thin convex lens of focal length 5 cm and a thin concave lens of focal length 4 cm are combined together (without any gap) and this combination has magnification

m_{1}

when an object is placed 10 cm before the convex lens. Keeping the positions of convex lens and object undisturbed a gap of 1 cm is introduced between the lenses by moving the concave lens away, which lead to a change in magnification of total lens system to

m_{2}

. The value of

\left| \frac{m_{1}}{m_{2}} \right|

\_\_\_\_

A virtual object has u > 0

When the first lens's image lies beyond the second lens, the object for the second lens is on its outgoing side. Taking u as negative there puts the final image on the wrong side.

Measure from the next lens

Before using the second lens, subtract the separation. The first image's distance from lens 1 is not its distance from lens 2.

Powers add only in contact

For lenses a distance d apart, P = P₁ + P₂ − dP₁P₂. Adding the powers alone ignores the gap.

Drill 13 more on combinations of lenses

Summary — formulas & gotchas at a glance

A revision cheat-sheet for the formulas and gotchas above. Click any concept name to jump back to its full explanation.

Formulas (2)

Thin lens formula and magnification
Thin lens formula
$\frac{1}{v} - \frac{1}{u} = \frac{1}{f}, \qquad m = \frac{v}{u} = \frac{f}{f + u}, \qquad P = \frac{1}{f}$
Combinations of lenses
Lenses in combination
$P = P_1 + P_2 - dP_1P_2, \qquad m = m_1 m_2$

Watch out for (6)

The lens formula has a minus sign
→ Thin lens formula and magnification
A concave lens never forms a real image of a real object
→ Thin lens formula and magnification
A long or slanted object needs two magnifications
→ Thin lens formula and magnification
A virtual object has u > 0
→ Combinations of lenses
Measure from the next lens
→ Combinations of lenses
Powers add only in contact
→ Combinations of lenses

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