12 min readConcept sheetSection A MCQ (1 mark) and fill-in-the-blank parts on power and images; Section B short answer (3 marks) numerical on the lens formula, or long answer (5 marks), often with ray diagrams; also a NEET favourite

Lenses: images, lens formula, magnification and power

A convex lens converges light and a concave lens diverges it. Use 1/f = 1/v − 1/u, m = v/u and power P = 1/f in dioptres.

Lens formula1/f = 1/v − 1/u

Magnificationm = size of image (I) ÷ size of object (O) = v/u

PowerP = 1/f, with f in metres; unit dioptre (D)

SignsConvex lens: f and P positive. Concave lens: f and P negative

Lens words and the two rays for drawing

A lens is a piece of transparent material bounded by two surfaces, at least one of them curved.
Kinds of lens: double convex, double concave, plano-convex, plano-concave, convexo-concave and concavo-convex.
A convex lens bulges outwards. It makes parallel rays converge (meet) at its focus, so it is a converging lens.
A concave lens caves inwards. It makes parallel rays diverge (spread) as if from its focus, so it is a diverging lens.
The optical centre O is the centre point of the lens. A lens has two foci, F₁ and F₂, one on each side; 2F marks twice the focal length.
Ray 1: a ray parallel to the axis passes through F₂ after a convex lens, or seems to come from F₁ after a concave lens.
Ray 2: a ray through the optical centre goes straight on without bending. Any two rays fix the image.

Images formed by lenses

Lens and object position	Image at	Size	Nature
Convex: at infinity	F₂	Highly diminished, point-sized	Real and inverted
Convex: beyond 2F₁	Between F₂ and 2F₂	Diminished	Real and inverted
Convex: at 2F₁	2F₂	Same size	Real and inverted
Convex: between F₁ and 2F₁	Beyond 2F₂	Enlarged	Real and inverted
Convex: at F₁	Infinity	Highly enlarged	Real and inverted
Convex: between F₁ and O	Same side as the object	Enlarged	Virtual and erect
Concave: at infinity	F₁	Highly diminished, point-sized	Virtual and erect
Concave: between infinity and O	Between F₁ and O	Diminished	Virtual and erect

How to draw the ray diagrams for convex and concave lenses

Draw two rays from the top point A in every case. One is parallel to the axis and bends through F₂. The other goes straight through O. Where they cross is A′; draw A′B′ to the axis. Mark O, F₁ and 2F₁ on the left and F₂ and 2F₂ on the right, evenly spaced.

1
Object at infinity. Draw two rays coming from the left, parallel to the axis. After the lens, draw both passing through F₂.
Image at F₂: point-sized, real, inverted.
2
Object beyond 2F₁. Stand AB to the left of 2F₁. The ray parallel to the axis bends through F₂. The ray through O goes straight. They cross below the axis between F₂ and 2F₂.
Image between F₂ and 2F₂: diminished, real, inverted.
3
Object at 2F₁. Stand AB on 2F₁. The same two rays cross below the axis exactly at 2F₂. Draw A′B′ pointing down, as long as AB.
Image at 2F₂: same size, real, inverted.
4
Object between F₁ and 2F₁. Stand AB between F₁ and 2F₁. The two rays cross below the axis beyond 2F₂. Draw a long A′B′ pointing down.
Image beyond 2F₂: enlarged, real, inverted.
5
Object at F₁. Stand AB on F₁. The ray parallel to the axis bends through F₂. The ray through O goes straight. The two refracted rays are parallel, so they meet only at infinity.
Image at infinity: highly enlarged, real, inverted.
6
Object between F₁ and O. Stand AB between F₁ and the lens. The ray parallel to the axis bends through F₂. The ray through O goes straight. The rays spread apart on the right. Extend both backwards as dotted lines: they meet at A′ above the axis, on the object's side. Draw A′B′ upright, larger and dotted.
Image on the object's side: enlarged, virtual, erect. This is how a magnifying glass works.
7
Concave lens, object at infinity. Draw two rays parallel to the axis. After the lens, draw them spreading outwards. Extend each backwards as a dotted line: they meet at F₁ on the left.
Image at F₁: point-sized, virtual, erect.
8
Concave lens, object anywhere in front. From A, draw a ray parallel to the axis. After the lens, bend it outwards and dot it back to F₁. From A, draw a ray through O; it goes straight. The dotted line and this ray meet at A′ between F₁ and O. Draw A′B′ upright, small and dotted.
Image between F₁ and O: diminished, virtual, erect.

Lens formula, magnification and power

Lens formula1/f = 1/v − 1/u. Measure all distances from the optical centre, with the same sign convention as mirrors.

Sign of fConvex lens: positive. Concave lens: negative.

Magnificationm = I/O = v/u. Negative m: real and inverted. Positive m: virtual and erect. For a lens there is no minus sign; for a mirror, m = −v/u.

Power PP = 1/f, with f in metres. A shorter focal length means a more powerful lens.

Unit of powerThe dioptre (D), spelt diopter in your lesson. 1 D is the power of a lens of focal length 1 m.

Sign of powerConvex lens: positive (+). Concave lens: negative (−).

Convex and concave lens numericals

Problem

(a) An object 2 cm tall is placed 30 cm from a convex lens of focal length 20 cm. Find the position, size and nature of the image, and the power of the lens. (b) An object 4 cm tall is placed 20 cm from a concave lens of focal length 20 cm. Find the image.

Working

(a) Given: u = −30 cm, f = +20 cm (convex). 1/v = 1/f + 1/u = 1/20 − 1/30.
LCM is 60: 1/v = (3 − 2)/60 = 1/60, so v = +60 cm.
m = v/u = 60/(−30) = −2, so I = −2 × 2 cm = −4 cm (inverted).
Power: f = 0.20 m, so P = 1/0.20 = +5 D.
(b) Given: u = −20 cm, f = −20 cm (concave). 1/v = 1/f + 1/u = −1/20 − 1/20 = −2/20, so v = −10 cm.
m = v/u = (−10)/(−20) = +0.5, so I = 0.5 × 4 cm = 2 cm (erect). Power = 1/(−0.20) = −5 D.

Answer

(a) Image 60 cm on the far side: real, inverted, 4 cm tall (twice the object); P = +5 D. (b) Image 10 cm from the lens on the object's side: virtual, erect, 2 cm tall; P = −5 D.

Activity (practical): image by a convex lens with a candle

You need a convex lens (focal length about 15 cm) in a holder and a candle on a stand. You also need a white screen, a metre scale and a matchbox.

1
Find a rough focal length first: focus a far-off tree or building on a white card and measure the lens-to-card distance.
2
Fix a metre scale on a table. Stand the candle, the convex lens and a screen on it, with the flame, lens centre and screen centre at the same height.
3
Place the lens a little more than one focal length from the candle. Move the screen until the flame's image is sharp.
4
Record the positions and work out the object distance and image distance. Note whether the image is real or virtual, erect or inverted, larger or smaller.
5
Move the candle further from the lens step by step, refocus, and record each time.
6
Now place the candle closer than the focal length. Try to catch the image on the screen.
No image forms on the screen. Looking through the lens, you see an enlarged, erect, virtual image.
7
Conclude.
As the object moves away (beyond F), the real image comes closer to the lens and gets smaller. At 2F it is the same size.

In real life

An optician's prescription such as +2.0 D or −1.5 D gives the power of each spectacle lens: plus is convex, minus is concave. A jeweller's magnifying glass is a convex lens held so that the gold sits inside F.

Exam angle

1 marks: “Define 1 dioptre.” 1 dioptre is the power of a lens whose focal length is 1 metre.

2 marks: “A lens has a power of −2.5 D. Name the lens and find its focal length.” Negative power means a concave lens. f = 1/P = 1/(−2.5) = −0.4 m = −40 cm.

3 marks: “An object is placed 20 cm from a convex lens of focal length 15 cm. Find the position and magnification of the image and state its nature.” u = −20 cm, f = +15 cm. 1/v = 1/15 − 1/20 = (4 − 3)/60 = 1/60, so v = +60 cm. m = v/u = 60/(−20) = −3. Real, inverted, three times larger, on the far side of the lens.

Remember it

A lens has a line (minus) in its formula: 1/v − 1/u. A mirror adds up: 1/v + 1/u. Power needs metres: change cm to m before you find 1/f.