Numford
Ray diagrams

Ray diagram of a concave lens: all 2 cases

A concave lens always forms a virtual, erect and diminished image on the same side as the object, between the focus F₁ and the optical centre O.

The ray diagram

Drag the object or type values. Signs follow the New Cartesian convention: distances are measured from the pole or optical centre, along the incident light is positive, against it negative, and heights above the axis are positive. So u is negative for a real object.

Image distance v-6.67 cmon the object side
Magnification m+0.33erect, 0.33 times
Image height h′+0.67 cmobject +2 cm
Power P-10 Ddiverging
NatureVirtual, erect, diminishedObject between infinity and the optical centre O
Image positionBetween F₁ and ODiminished

Concave lens of focal length -10 cm. An object +2 cm tall at u = -20 cm: the image is at v = -6.67 cm, virtual, erect, diminished.

Unit
cm
Show the working
  1. f = -10 cm, u = -20 cm (the object is in front, so u is negative), h = +2 cm.
  2. 1/v = 1/f + 1/u = 1/(-10) + 1/(-20) = -30/200 = -3/20
  3. v = -6.67 cm: negative, so the image is on the object side and virtual.
  4. m = v/u = (-6.67)/(-20) = +0.33: positive, so erect, and diminished.
  5. h′ = m × h = +0.33 × +2 = +0.67 cm
  6. Power P = 1/f = 1/(-0.1 m) = -10 D
All the cases

2 positions of the object

NCERT table

Image formation by a concave lens

ObjectImageSizeNature
At infinityAt the focus F₁Highly diminished, point-sizedVirtual and erect
Between infinity and the optical centre OBetween F₁ and ODiminishedVirtual and erect
How to draw it

The rays to use

  • A ray parallel to the principal axis spreads out after refraction, as if it came from the focus F₁ on the object side.
  • A ray through the optical centre O goes straight on without bending.
  • A ray heading towards the focus F₂ on the far side comes out parallel to the principal axis.
  • The rays spread out, so the image is where their extensions meet, on the object side.
With numbers

Object between infinity and O, f = 10 cm

  1. f = -10 cm, u = -20 cm (the object is in front, so u is negative), h = +2 cm.
  2. 1/v = 1/f + 1/u = 1/(-10) + 1/(-20) = -30/200 = -3/20
  3. v = -6.67 cm: negative, so the image is on the object side and virtual.
  4. m = v/u = (-6.67)/(-20) = +0.33: positive, so erect, and diminished.
  5. h′ = m × h = +0.33 × +2 = +0.67 cm
  6. Power P = 1/f = 1/(-0.1 m) = -10 D
Questions

Quick answers

Can a concave lens form a real image?

No, not of a real object. The refracted rays always spread out, so they only appear to come from a point on the object side.

Where is a concave lens used?

In spectacles for short sight (myopia), and in door peepholes and some telescopes.

Where is the image of a far object?

At the focus F₁ on the object side, highly diminished and point-sized.

Try your own mirror or lens